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283 Find the unit digit of $\left(\frac{5+\sqrt{21}}{2}\right)^{2010}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
|
Solve the problem, which is to find the remainder when $\left[\left(\frac{5+\sqrt{21}}{2}\right)^{2010}\right]$ is divided by 10.
Let $a_{n}=\left(\frac{5+\sqrt{21}}{2}\right)^{n}+\left(\frac{5-\sqrt{21}}{2}\right)^{n}\left(n \in \mathbf{N}_{+}\right)$.
Since $\frac{5+\sqrt{21}}{2}+\frac{5-\sqrt{21}}{2}=5$,
$$
\frac{5+\sqrt{21}}{2} \cdot \frac{5-\sqrt{21}}{2}=1 \text {, }
$$
Therefore, $\frac{5+\sqrt{21}}{2} ， \frac{5-\sqrt{21}}{2}$ are the roots of the equation $x^{2}-5 x+1=0$.
Thus, $a_{n}-5 a_{n-1}+a_{n-2}=0$, which means
$$
a_{n}=5 a_{n-1}-a_{n-2}(n \geqslant 3) \text {. }
$$
Also, $a_{1}=5, a_{2}=23$ are both integers, and from equation (1), we know that $a_{n}$ is an integer for all positive integers $n$.
Hence, $a_{2010}$ is an integer.
Since $4<\sqrt{21}<5$, then
$0<5-\sqrt{21}<1$.
Therefore, $0<\frac{5-\sqrt{21}}{2}<1$.
Thus, $0<\left(\frac{5-\sqrt{21}}{2}\right)^{2010}<1$.
Consequently, $a_{2010}$ is the smallest integer greater than $\left(\frac{5+\sqrt{21}}{2}\right)^{2010}$.
Therefore, $\left[\left(\frac{5+\sqrt{21}}{2}\right)^{2010}\right]=a_{2010}-1$.
From equation (1), we have
$$
\begin{array}{l}
a_{n} \equiv 5 a_{n-1}-a_{n-2} \equiv-a_{n-2}(\bmod 5) \\
\Rightarrow a_{n} \equiv-a_{n-2} \equiv-\left(-a_{n-4}\right) \equiv a_{n-4}(\bmod 5) \\
\Rightarrow a_{2010} \equiv a_{2} \equiv 23 \equiv 3(\bmod 5) .
\end{array}
$$
From equation (1), we also have
$$
\begin{aligned}
a_{n} & \equiv a_{n-1}+a_{n-2}(\bmod 2) \\
\Rightarrow & a_{n} \equiv a_{n-1}+a_{n-2} \\
& \equiv\left(a_{n-2}+a_{n-3}\right)+a_{n-2} \\
& \equiv 2 a_{n-2}+a_{n-3} \equiv a_{n-3}(\bmod 2) \\
\Rightarrow & a_{2010} \equiv a_{3} \equiv a_{2}+a_{1} \equiv 0(\bmod 2) .
\end{aligned}
$$
Thus, $a_{2010} \equiv 3(\bmod 5)$, and
$$
a_{2010} \equiv 0(\bmod 2) \text {. }
$$
Therefore, $a_{2010} \equiv 8(\bmod 10)$.
Hence, $\left[\left(\frac{5+\sqrt{21}}{2}\right)^{2010}\right]=a_{2010}-1$
$$
\equiv 7(\bmod 10) \text {. }
$$
Thus, 7 is the answer.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 On a plane, there are seven points, and some line segments can be connected between them, so that any three points among the seven must have at least one line segment connecting a pair of them. How many line segments are needed at least? Prove your conclusion.
$(2002$, Shanghai Junior High School Mathematics Competition)
|
(1) If point $A$ is not an endpoint of any line segment, then every two of the other six points are connected by a line segment, totaling $\frac{1}{2} \times 6 \times 5=15$ line segments;
(2) If point $A$ is the endpoint of only one line segment, then every two of the five points not connected to $A$ are connected by a line segment, at least $\frac{1}{2} \times 5 \times 4=10$ line segments, making a total of 11 line segments;
(3) If point $A$ is the endpoint of only two line segments $A B$ and $A C$, then every two of the four points not connected to $A$ are connected by a line segment, at least $\frac{1}{2} \times 4 \times 3=6$ line segments. One more line segment must be drawn from point $B$, so at least $2+6+1=9$ line segments are needed.
(4) If each point is the endpoint of at least three line segments, then at least $\frac{1}{2} \times 7 \times 3=\frac{21}{2}$ line segments are needed. Since the number of line segments must be an integer, at least 11 line segments are needed.
Figure 2 shows that nine line segments are sufficient.
|
9
|
Combinatorics
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given 8 points $A_{1}, A_{2}, \cdots, A_{8}$ on a circle. Find the smallest positive integer $n$, such that among any $n$ triangles with these 8 points as vertices, there must be two triangles that share a common side.
(Tao Pingsheng, problem contributor)
|
8. First, consider the maximum number of triangles with no common edges.
Connecting every pair of eight points yields $\mathrm{C}_{8}^{2}=28$ chords.
If each chord belongs to only one triangle, then these chords can form at most $r \leqslant\left[\frac{28}{3}\right]=9$ triangles with no common edges. However, if there are nine such triangles, they would have a total of 27 vertices, meaning that one of the eight points (let's say $A_{8}$) must belong to at least four triangles. The opposite sides of the four triangles sharing vertex $A_{8}$ are all lines connecting two points among $A_{1}, A_{2}, \cdots, A_{7}$. Therefore, one point (let's say $A_{k}$) must appear twice, which means the corresponding two triangles will share a common edge $A_{8} A_{k}$, leading to a contradiction. Therefore, $r \leqslant 8$.
On the other hand, it is possible to construct eight triangles such that any two of them have no common edges.
Notice that such eight triangles produce a total of 24 vertices, and each point belongs to at most three triangles. Hence, each point belongs to exactly three triangles, meaning each point has a degree of 6.
For simplicity, as shown in Figure 3, take the eight equally spaced points on the circumference as the eight vertices of graph $G$, construct an 8-vertex complete graph, and then remove four diameters. Thus, we get 24 edges, and each point belongs to six edges. Among the triangles formed by these edges, select eight isosceles triangles:
$$
(1,2,3),(3,4,5),(5,6,7),(7,8,1)
$$
and $(1,4,6),(3,6,8),(5,8,2),(7,2,4)$, which have no common edges (each set of four triangles can be obtained by rotating one of the triangles around the center appropriately).
Therefore, $r=8$.
Thus, the required minimum value $n=r+1=9$.
|
9
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Robots A and B start from the starting point at the same time, moving uniformly along a hundred-meter track, and the automatic recorder shows: when A is $1 \mathrm{~m}$ away from the finish line, B is $2 \mathrm{~m}$ away from the finish line; when A reaches the finish line, B is $1.01 \mathrm{~m}$ away from the finish line. Then the actual length of this track is more than $100 \mathrm{~m}$ by $\qquad$
|
2.1.
Let the actual length of the track be $x \mathrm{~m}$, and the speeds of robots 甲 and 乙 be $v_{\text {甲 }}$ and $v_{\text {乙}}$, respectively. Thus,
$$
\frac{v_{\text {甲 }}}{v_{\text {乙 }}}=\frac{x-1}{x-2}=\frac{x}{x-1.01} \text {. }
$$
Solving for $x$ yields $x=101$.
Therefore, this track is $1 \mathrm{~m}$ longer than $100 \mathrm{~m}$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Among the seven points consisting of the center and the six vertices of a regular hexagon, if any $n$ points are taken, and among them, there must be three points that form the vertices of an equilateral triangle, then the minimum value of $n$ is $\qquad$
|
-1.5 .
Consider the regular hexagon $A B C D E F$ as shown in Figure 1, with its center at $O$.
When $n=4$, take $A, C, D, F$, among which no three points can form the three vertices of an equilateral triangle.
When $n=5$, consider the following two cases:
(1) $O$ is among the five points. Consider the three pairs of points $(A, B)$, $(C, D)$, $(E, F)$. According to the pigeonhole principle, at least one pair of points must be taken, which, together with the center $O$, can form the three vertices of an equilateral triangle.
(2) $O$ is not among the five points. Consider the two groups of points $(A, C, E)$ and $(B, D, F)$. According to the pigeonhole principle, at least one group of points must be taken, which can form the three vertices of an equilateral triangle.
In summary, the minimum value of $n$ is 5.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. As shown in Figure 1, in the cube $A C_{1}$ with edge length 1, points $P$ and $Q$ are moving points on edges $A D$ and $A_{1} B_{1}$, respectively. If the skew lines $B D_{1}$ and $P Q$ are perpendicular to each other, then $A P + A_{1} Q =$ $\qquad$
|
5. 1 .
As shown in Figure 5, establish a spatial rectangular coordinate system, and set
$$
\begin{array}{l}
B(1,0,0), \\
D_{1}(0,1,1), \\
P(0, a, 0), \\
Q(b, 0,1) . \\
\text { Then } \overrightarrow{B D_{1}}=(-1,1,1), \\
\overrightarrow{P Q}=(b,-a, 1) .
\end{array}
$$
$$
\text { Therefore, } 0=\overrightarrow{B D_{1}} \cdot \overrightarrow{P Q}=-b-a+1 \text {. }
$$
Thus, $a+b=1$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. For a positive integer $x$, let $S(x)$ denote the sum of the digits of $x$. Then the maximum value of $S(x)-9[\lg x]$ is $\qquad$ ( $[x]$ denotes the greatest integer not exceeding the real number $x$).
|
8. 9 .
Let $x=\overline{a_{n} a_{n-1} \cdots a_{0}}\left(a_{0} \neq 0\right)$. Then
$$
\begin{array}{l}
10^{n} \leqslant x<10^{n+1} \\
\Rightarrow n \leqslant \lg x<n+1 \Rightarrow[\lg x]=n . \\
\text { Hence } S(x)=a_{0}+a_{1}+\cdots+a_{n} \leqslant 9(n+1) \\
=9([\lg x]+1) .
\end{array}
$$
Therefore, $S(x)-9[\lg x] \leqslant 9$.
$$
\text { Also, } S(9)-9[\lg 9]=9 \text {, then, } S(x)-9[\lg x]
$$
has a maximum value of 9.
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Let $n$ be a positive integer, and $f(n)$ denote the number of integers satisfying the following conditions:
(i) Each digit $a_{i} \in\{1,2,3,4\}$, and
$a_{i} \neq a_{i+1}(i=1,2, \cdots)$;
(ii) When $n \geqslant 3$, $a_{i}-a_{i+1}$ and $a_{i+1}-a_{i+2}$ $(i=1,2, \cdots)$ have opposite signs.
(1) Find the value of $f(10)$;
(2) Determine the remainder when $f(2008)$ is divided by 13.
[3]
|
Let $g(n)$ denote the number of wave numbers for which $a_{n}>a_{n-1}$ when $n \geqslant 2$. Then, by symmetry, we have
$$
\begin{array}{l}
g(n)=\frac{1}{2} f(n) . \\
\text { Hence } a_{n-1}=1, a_{n}=2,3,4 ; \\
a_{n-1}=2, a_{n}=3,4 ; \\
a_{n-1}=3, a_{n}=4 .
\end{array}
$$
Let $m(i)$ represent the number of $(n-1)$-digit wave numbers where $a_{n-2}>a_{n-1}$ and the last digit is $i$. Then
$$
\begin{array}{l}
m(1)+m(2)+m(3)+m(4)=g(n-1), \\
m(4)=0 .
\end{array}
$$
When $a_{n-1}=1$, $a_{n-2}=2,3,4$. Then
$$
m(1)=g(n-2) \text {. }
$$
When $a_{n-1}=3$, $a_{n-2}=4, a_{n-3}=1,2,3$. Then $m(3)=g(n-3)$.
Thus, $g(n)=3 m(1)+2 m(2)+m(3)$
$$
=2 g(n-1)+g(n-2)-g(n-3) \text {. }
$$
It is easy to see that $g(2)=6, g(3)=14, g(4)=31$, $\cdots \cdots g(10)=4004$.
Therefore, $f(10)=2 g(10)=8008$.
Also note that the sequence of remainders when $g(n)$ is divided by 13 is:
$$
6,1,5,5,1,2,0,1,0,1,1,3,6,1,5,5, \cdots
$$
The smallest positive period is 12.
Thus, $g(2008) \equiv g(4) \equiv 5(\bmod 13)$.
Therefore, $f(2008) \equiv 10(\bmod 13)$.
|
10
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Let $f(x)$ represent a fourth-degree polynomial in $x$. If
$$
f(1)=f(2)=f(3)=0, f(4)=6, f(5)=72 \text {, }
$$
then the last digit of $f(2010)$ is $\qquad$
|
6. 2 .
Since $f(1)=f(2)=f(3)=0$, the quartic polynomial can be set as
$$
\begin{array}{l}
f(x)=(x-1)(x-2)(x-3)(a x+b) . \\
\text { By } f(4)=6, f(5)=72, \text { we get } \\
\left\{\begin{array} { l }
{ 6 ( 4 a + b ) = 6 , } \\
{ 2 4 ( 5 a + b ) = 7 2 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a=2, \\
b=-7 .
\end{array}\right.\right.
\end{array}
$$
Then $f(2010)$
$$
\begin{array}{l}
=(2010-1)(2010-2)(2010-3)(2010 \times 2-7) \\
=2009 \times 2008 \times 2007 \times 4013 .
\end{array}
$$
Therefore, the last digit of $f(2010)$ is 2.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Let $p, q$ be prime numbers, and satisfy $p^{3}+q^{3}+1=p^{2} q^{2}$. Then the maximum value of $p+q$ is
|
8.5.
Assume $q \leqslant p$. Notice that
$$
\begin{array}{l}
p^{3}+q^{3}+1=p^{2} q^{2} \\
\Rightarrow q^{3}+1=p^{2} q^{2}-p^{3} \\
\Rightarrow(q+1)\left(q^{2}-q+1\right)=p^{2}\left(q^{2}-p\right) .
\end{array}
$$
Therefore, $p^{2} \mid(q+1)\left(q^{2}-q+1\right)$.
Since $q \leqslant p$, we have $0<q^{2}-q+1<p^{2}$.
This implies $p \mid(q+1)$.
Also, since $q \leqslant p$, we have $q+1=p$.
Given that $p$ and $q$ are both prime numbers, the only solution is $p=3$ and $q=2$. Thus, the maximum value of $p+q$ is 5.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Line segment $A B$ divides a square into two polygons (points $A$ and $B$ are on the sides of the square), each polygon has an inscribed circle, one of which has a radius of 6, while the other has a radius greater than 6. What is the difference between the side length of the square and twice the length of line segment $A B$?
|
If line segment $A B$ divides the square into two triangles, then $A B$ can only be the diagonal of the square. But in this case, the radii of the two incircles are equal, which contradicts the condition.
If one of the polygons is a quadrilateral, then the sum of the lengths of $A B$ and its opposite side is greater than the sum of the lengths of the other two sides, which contradicts the fact that the quadrilateral has an incircle.
Therefore, it can only be that $A B$ divides the square into a triangle and a pentagon, and the incircle of the pentagon is also the incircle of the original square (as shown in Figure 9).
As shown in Figure 10, let the incircle of $\triangle A B C$ touch $B C$, $C A$, and $A B$ at points $D$, $E$, and $F$, respectively, and the incircle of the original square, which is also the excircle of $\triangle A B C$, touch $C A$, $B C$, and $A B$ at points $P$, $Q$, and $R$, respectively. Then the side length of the original square is $C P + C Q$.
$$
\begin{aligned}
\text { Hence } & C P + C Q - 2 A B \\
= & (A P + A E + 6) + (B Q + B D + 6) - \\
& (A F + B F + A R + B R) \\
= & 12 + (A P - A R) + (B Q - B R) + \\
& (A E - A F) + (B D - B F) \\
= & 12 .
\end{aligned}
$$
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. (40 points) Let integers $m, n$ satisfy $m \geqslant n$ and $m^{3}+n^{3}+1=4 m n$. Find the maximum value of $m-n$.
保留了原文的换行和格式，如上所示。
|
3. Let $s=m+n, p=mn$. Then $m^{3}+n^{3}=(m+n)^{3}-3m^{2}n-3mn^{2}$ $=s^{3}-3ps$.
From $s^{3}-3ps+1=4p$, we get $p=\frac{s^{3}+1}{3s+4}$.
Since $m, n$ are integers, 27p must also be an integer, and
$27p=9s^{2}-12s+16-\frac{37}{3s+4}$.
Therefore, $3s+4$ must divide 37.
Thus, $3s+4 = \pm 1$ or $\pm 37$.
If $3s+4=-1$ or -37, then $s$ is not an integer;
If $3s+4=37$, then $s=11$ and $p=36$, but there are no integers $m, n$;
If $3s+4=1$, then $s=-1$ and $p=0$.
Thus, $m=0, n=-1$.
Therefore, the maximum value of $m-n$ is 1.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. (40 points) Write "KOREAIMC" in eight lines as shown in Figure 5, where the first line has one K, the second line has two O's, and so on, with the last line having eight C's.
$$
\begin{array}{llllllll}
\mathbf{K} & & & & & & & \\
\mathbf{O} & \mathbf{O} & & & & & & \\
\mathbf{R} & \mathbf{R} & \mathbf{R} & & & & & \\
\mathrm{E} & \mathbf{E} & \mathrm{E} & \mathrm{E} & & & & \\
\mathrm{A} & \mathrm{A} & \mathbf{A} & \mathrm{A} & \mathrm{A} & & & \\
\mathrm{I} & \mathrm{I} & \mathrm{I} & \mathbf{I} & \mathrm{I} & \mathrm{I} & & \\
\mathrm{M} & \mathrm{M} & \mathbf{M} & \mathrm{M} & \mathrm{M} & \mathrm{M} & \mathrm{M} & \\
\mathrm{C} & \mathrm{C} & \mathrm{C} & \mathrm{C} & \mathrm{C} & \mathrm{C} & \mathrm{C} & \mathrm{C}
\end{array}
$$
Figure 5
Starting from the K at the top, move downward row by row to spell out "KOREAIMC". Each move can only be to the letter directly below or to a letter adjacent to the one directly below. The bold letters in Figure 5 represent one such path. If a letter is removed from Figure 5, such that the number of different remaining paths is only 516, identify the position of the removed letter.
|
10. First list the number table as shown in Figure 17, where each cell corresponds to a letter, and the number in the cell represents the number of different paths from the letter "K" to that letter. The cell at the very top is filled with 1. Starting from the second row, the number in each cell is the sum of the numbers in the cells from the row above that share a common vertex with it.
Then list
the number table as shown in Figure 18,
where each cell
also corresponds
to a letter, and the
number in the cell
indicates the number of different paths from that letter to the last row, filled in a similar manner to the table in Figure 17.
Thus, for each letter, the number table in Figure 17 gives the number of paths \( m \) from the letter "K" to that letter, while the number table in Figure 18 gives the number of paths \( n \) from that letter to the last row. If the letter is removed, the number of paths will be reduced by \( m \times n \). If the letter is not removed, the total number of different paths is 750. Since there are now only 516 paths, the total reduction is
$$
750 - 516 = 234 = 2 \times 3^{2} \times 13
$$
paths.
It is easy to see that only the numbers 13 and 26 in the fifth row of the number table in Figure 18 are multiples of 13.
At this point, \( 13 \times 9 = 117, 26 \times 9 = 234 \).
Therefore, the letter removed is the third letter "A" in the fifth row of Figure 5.
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let $k$ be an integer greater than 1, and the sequence $\left\{a_{n}\right\}$ is defined as follows:
$$
\begin{array}{l}
a_{0}=0, a_{1}=1, \\
a_{n+1}=k a_{n}+a_{n-1}(n=1,2, \cdots) .
\end{array}
$$
Find all $k$ that satisfy the following condition: there exist non-negative integers $l, m (l \neq m)$, and positive integers $p, q$, such that
$a_{l}+k a_{0}=a_{m}+k a_{q^{*}} \quad$ (Xiong Bin)
|
When $k=2$, $a_{0}=0, a_{1}=1, a_{2}=2$, then from $a_{0}+2 a_{2}=a_{2}+2 a_{1}=4$, we know that taking $l=0, m=2, p=$ $2, q=1$ is sufficient.
For $k \geqslant 3$, by the recurrence relation, $\left\{a_{n}\right\}$ is a strictly increasing sequence of natural numbers and $k \mid\left(a_{n+1}-a_{n-1}\right)$.
$$
\begin{array}{l}
\text { then } a_{2 n} \equiv a_{0}=0(\bmod k), \\
a_{2 n+1} \equiv a_{1}=1(\bmod k)(n=0,1, \cdots) .
\end{array}
$$
If there exist $l, m \in \mathbf{N}, p, q \in \mathbf{N}_{+}, l \neq m$, satisfying $a_{l}+k a_{p}=a_{m}+k a_{q}$, without loss of generality, assume $l0$, then
$$
\begin{array}{l}
a_{l}+k a_{p} \leqslant k a_{p}+a_{p-1}=a_{p+1} \\
\leqslant a_{m}\frac{a_{l}+k a_{p}-a_{m}}{k}=a_{q} \text {. }
\end{array}
$$
From $k a_{q}+a_{m}=k a_{p}+a_{l} \geqslant k a_{p}$,
we know $a_{q} \geqslant a_{p}-\frac{a_{m}}{k} \geqslant a_{p}-\frac{a_{p}}{k}=\frac{k-1}{k} a_{p}$.
Noting that $a_{p} \geqslant k a_{p-1}$. Thus,
$a_{p}>a_{q} \geqslant \frac{k-1}{k} a_{p} \geqslant(k-1) a_{p-1} \geqslant a_{p-1}$.
From equation (4), we know $a_{q}=a_{p-1}$.
Therefore, the equalities in equations (1), (2), and (3) must all hold.
From equations (2) and (3), we get
$m=p, p=2$.
Thus, $a_{q}=a_{p-1}=a_{1}=1$.
From equation (1), we know $l=0$.
Therefore, from $a_{l}+k a_{p}=a_{m}+k a_{q}$, we get $k^{2}=k+k$, i.e., $k=2$, which is a contradiction.
Thus, $k \geqslant 3$ does not satisfy the problem's conditions.
Hence, $k$ can only be 2.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Find the value of $\frac{\log _{2} \frac{2}{3}+\log _{2} \frac{3}{4}+\log _{2} \frac{4}{5}+\log _{2} \frac{5}{6}+\log _{2} \frac{6}{7}+\log _{2} \frac{7}{8}}{\log _{3} 3 \cdot \log _{4} 3 \cdot \log _{5} 4 \cdot \log _{8} 5 \cdot \log _{7} 6 \cdot \log _{8} 7}$.
|
2. -6 .
$$
\begin{array}{l}
\text { Original expression }=\frac{\log _{2}\left(\frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} \cdot \frac{5}{6} \cdot \frac{6}{7} \cdot \frac{7}{8}\right)}{\frac{\lg 2}{\lg 3} \cdot \frac{\lg 3}{\lg 4} \cdot \frac{\lg 4}{\lg 5} \cdot \frac{\lg 5}{\lg 6} \cdot \frac{\lg 6}{\lg 7} \cdot \frac{\lg 7}{\lg 8}} \\
=\frac{\log _{2} \frac{2}{8}}{\frac{\lg 2}{\lg 8}}=\frac{\log _{2} 2^{-2}}{\frac{\lg 2}{3 \lg 2}}=\frac{-2}{\frac{1}{3}}=-6 .
\end{array}
$$
|
-6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given that $x_{1}, x_{2}, \cdots, x_{2010}$ are all positive real numbers. Then
$$
x_{1}+\frac{x_{2}}{x_{1}}+\frac{x_{3}}{x_{1} x_{2}}+\cdots+\frac{x_{2010}}{x_{1} x_{2} \cdots x_{200}}+\frac{4}{x_{1} x_{2} \cdots x_{2010}}
$$
the minimum value is $\qquad$
|
3.4.
Starting from the last two terms, repeatedly applying the AM-GM inequality, we get
$$
\begin{array}{l}
\text { Original expression }=\sum_{i=1}^{2010} \frac{x_{i}}{\prod_{j=1}^{i-1} x_{j}}+\frac{4}{\prod_{j=1}^{2010} x_{j}} \\
=\sum_{i=1}^{2009} \frac{x_{i}}{\prod_{j=1}^{i-1} x_{j}}+\left(\frac{x_{2010}}{\prod_{j=1}^{2009} x_{j}}+\frac{4}{x_{2010} \prod_{j=1}^{2009} x_{j}}\right) \\
\geqslant \sum_{i=1}^{2009} \frac{x_{i}}{\prod_{j=1}^{i-1} x_{j}}+\frac{4}{\prod_{j=1}^{2009} x_{j}} \geqslant \cdots \geqslant x_{1}+\frac{4}{x_{1}} \geqslant 4,
\end{array}
$$
where the equality holds if and only if
$$
x_{2010}=x_{2009}=\cdots=x_{1}=2 \text {. }
$$
Therefore, the minimum value sought is 4.
|
4
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given that the circumcenter, incenter, and orthocenter of a non-isosceles acute $\triangle ABC$ are $O, I, H$ respectively, and $\angle A=60^{\circ}$. If the altitudes of $\triangle ABC$ are $AD, BE, CF$, then the ratio of the circumradius of $\triangle OIH$ to the circumradius of $\triangle DEF$ is $\qquad$ .
|
4. 2 .
From $\angle B O C=\angle B I C=\angle B H C=120^{\circ}$, we know that $O, I, H, B, C$ are concyclic.
Let the circumradii of $\triangle A B C$ and $\triangle O I H$ be $R$ and $r$, respectively.
By the Law of Sines, we have
$2 R \sin A=B C=2 r \sin \angle B O C$.
Thus, $r=R$.
Let the circumradius of $\triangle D E F$ be $r^{\prime}$.
Noting that $A, E, H, F$ are concyclic, and $A H$ is the diameter of this circle. By the Law of Sines, we get
$$
E F=A H \sin A=2 R \cos A \cdot \sin A=\frac{\sqrt{3}}{2} R .
$$
Since $A, F, D, C$ and $A, E, D, B$ are concyclic, respectively, we have
$\angle B D F=\angle C D E=\angle B A C=60^{\circ}$.
Thus, $\angle F D E=60^{\circ}$.
By the Law of Sines, we get
$$
E F=2 r^{\prime} \sin \angle F D E=\sqrt{3} r^{\prime} \text {. }
$$
Therefore, $r^{\prime}=\frac{R}{2}$.
Hence, $\frac{r}{r^{\prime}}=2$.
[Note] The circumcircle of $\triangle D E F$ is actually the nine-point circle of $\triangle A B C$, so $r^{\prime}=\frac{R}{2}$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. (15 points) In a regular pentagon $A B C D E$, the diagonal $B E$ intersects diagonals $A D$ and $A C$ at points $F$ and $G$, respectively. The diagonal $B D$ intersects diagonals $C A$ and $C E$ at points $H$ and $I$, respectively. The diagonal $C E$ intersects diagonal $A D$ at point $J$. Let the set of isosceles triangles formed by the 10 points $A, B, C, D, E, F, G, H, I, J$ and the line segments in Figure 2 be denoted as $M$.
(1) Determine the number of elements in $M$;
(2) If each of the 10 points is arbitrarily colored either red or blue, does there necessarily exist an isosceles triangle in $M$ whose three vertices are the same color?
(3) If $n$ of the 10 points are arbitrarily colored red such that there necessarily exists an isosceles triangle in $M$ whose three vertices are all red, find the minimum value of $n$.
|
11. (1) Since all triangles formed by 10 points $A, B, C, D, E, F, G, H, I, J$ and line segments in Figure 2 are isosceles triangles,
$$
|M|=\mathrm{C}_{5}^{3}+4 \mathrm{C}_{5}^{4}+5 \mathrm{C}_{5}^{5}=35 \text{. }
$$
(2) By the pigeonhole principle, we know that among $A, B, C, D, E$, there must be three points of the same color, and the triangle formed by these three points belongs to $M$. Therefore, there must exist an isosceles triangle in $M$ whose three vertices are of the same color.
(3) If $n=5$, then coloring $F, G, H, I, J$ red, there will be no triangle in $M$ whose vertices are all red.
If $n=6$, when there are at least three red points among $A, B, C, D, E$, there must exist a triangle in $M$ whose vertices are all red; when there are fewer than three red points among $A, B, C, D, E$, there are at least four red points among $F, G, H, I, J$.
If there are exactly four red points among $F, G, H, I, J$, assume $F, G, H, J$ are red points, then there are at least two red points among $A, B, C, D, E$. If one of $A, B, E$ is a red point, assume $A$ is a red point, then $\triangle AFG$ is a triangle in $M$ whose vertices are all red; otherwise, $C, D$ are both red, thus, $\triangle CDH$ is a triangle in $M$ whose vertices are all red.
If $F, G, H, I, J$ are all red, then there is at least one red point among $A, B, C, D, E$. Assume $A$ is a red point, then $\triangle AFG$ is a triangle in $M$ whose vertices are all red. Therefore, the minimum value of $n$ is 6.
(Li Jianquan provided)
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Divide a wire of length $11 \mathrm{~cm}$ into several segments of integer centimeters, such that any three segments can form the sides of a triangle. Then the number of different ways to divide the wire is $\qquad$ (ways with the same number of segments and corresponding equal lengths are considered the same way).
|
4.8.
Let the shortest two segments be $a \mathrm{~cm}, b \mathrm{~cm}$, and take any other segment $c \mathrm{~cm} (a \leqslant b \leqslant c)$. Then $b \leqslant c < a + b$.
(1) If $a = 1$, then $1 \leqslant b \leqslant c < b + 1$. Thus, $b = c$. Therefore, $b \in (11 - 1), 2b < 11 - 1$. Hence, $b$ can be $1, 2, 5$, giving three ways to divide.
(2) If $a = 2$, then $2 \leqslant b \leqslant c < b + 2$. Thus, $c$ can be $b$ or $b + 1$. Let the segments of length $b \mathrm{~cm}$ and $(b + 1) \mathrm{~cm}$ (excluding the segment of length $a$) be $x$ and $y$ segments respectively, where $x \geqslant 0$, $y \geqslant 0$, and $x + y \geqslant 2$.
By the problem, $2 + bx + (b + 1)y = 11$.
Then $y = \frac{9 + x}{b + 1} - x$.
When $x = 0$, $y = \frac{9}{b + 1}$, so $b + 1$ can only be 3, giving one way to divide $(2, 3, 3, 3)$;
When $x = 1$, $y = \frac{10}{b + 1} - 1$, so $b + 1$ can only be 5, giving one way to divide $(2, 4, 5)$;
When $x = 2$, it does not meet the problem's conditions;
When $x = 3$, $y = \frac{12}{b + 1} - 3$, so $b + 1 = 3, 4$, giving two ways to divide $(2, 2, 2, 2, 3), (2, 3, 3, 3)$;
When $x \geqslant 4$, $bx \leqslant 9 \Rightarrow b \leqslant \frac{9}{x} \leqslant \frac{9}{4} < 3$, thus $b = 2$, then $2 + 2x + 3y = 11$, solving gives
$$
(x, y) = (3, 1), (0, 3),
$$
giving two ways to divide $(2, 2, 2, 2, 3), (2, 3, 3, 3)$.
(3) If $a \geqslant 3$, then $a + b + c \geqslant 9, 2a + b + c \geqslant 12$. Therefore, the wire can only be divided into 3 segments. Thus, $(a, b, c) = (3, 3, 5), (3, 4, 4)$, giving two ways to divide.
In summary, there are 8 ways to divide in total.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. In space, there are five points, no four of which are coplanar. If several line segments are drawn such that no tetrahedron exists in the graph, then the maximum number of triangles in the graph is $\qquad$.
|
14.4.
First, construct graph 6. It is easy to see that it meets the conditions and has exactly four triangles.
Now assume there exists some configuration where the number of triangles is no less than five.
If only two line segments are not connected, then these two line segments must have no common endpoints (as shown in graph 6), otherwise, a tetrahedron would exist. But there are only four triangles, which is a contradiction.
If at least three line segments are not connected, when one of these line segments serves as a side of three triangles, as shown in graph 7, there are only three triangles; when each line segment serves as a side of at most two triangles, then there are at most $\left[\frac{\left(\mathrm{C}_{5}^{2}-3\right) \times 2}{3}\right]=4$ triangles.
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
18. (17 points) Let $a_{1}, a_{2}, \cdots, a_{n}$ be a permutation of the integers 1, 2, $\cdots, n$, and satisfy the following conditions:
(1) $a_{1}=1$;
(2) $\left|a_{i}-a_{i+1}\right| \leqslant 2(i=1,2, \cdots, n-1)$.
Let the number of such permutations be $f(n)$. Find the remainder when $f(2010)$ is divided by 3.
|
18. Verifiable
$$
f(1)=1, f(2)=1, f(3)=2 \text {. }
$$
Let $n \geqslant 4$. Then $a_{2}=2$ or 3.
For $a_{2}=2$, the number of permutations is $f(n-1)$. This is because by removing the first term and reducing all subsequent terms by 1, a one-to-one correspondence can be established.
For $a_{2}=3$, if $a_{3}=2$, then $a_{4}=4$, so the number of permutations is $f(n-3)$; if $a_{3} \neq 2$, then 2 must be placed after 4, leading to all odd numbers being in ascending order followed by all even numbers in descending order.
Therefore, $f(n)=f(n-1)+f(n-3)+1$.
Let $r(n)$ be the remainder when $f(n)$ is divided by 3. Then
$$
r(1)=r(2)=1, r(3)=2 \text {. }
$$
For $n \geqslant 4$,
$$
r(n) \equiv[r(n-1)+r(n-3)+1](\bmod 3) \text {. }
$$
Thus, $\{r(n)\}$ forms a sequence with a period of 8:
$$
1,1,2,1,0,0,2,0, \cdots \text {. }
$$
Since $2010 \equiv 2(\bmod 8)$, we have
$$
r(2010)=1,
$$
which means the remainder when $f(2010)$ is divided by 3 is 1.
|
1
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) Find the smallest positive integer $n$, such that among any $n$ irrational numbers, there are always three numbers, the sum of any two of which is irrational.
|
Three, take four irrational numbers $\{\sqrt{2}, \sqrt{3},-\sqrt{2},-\sqrt{3}\}$, obviously they do not satisfy the condition, hence $n \geqslant 5$.
Consider five irrational numbers $a, b, c, d, e$, viewed as five points.
If the sum of two numbers is a rational number, then connect the corresponding two points with a red line. Otherwise, connect them with a blue line.
(1) There is no red triangle. Otherwise, assume without loss of generality that $a+b$, $b+c$, and $c+a$ are all rational numbers. But $(a+b)+(c+a)-(b+c)=2a$, which contradicts the fact that $a$ is irrational.
(2) There must be a monochromatic triangle. Otherwise, there must be a red circle in the graph, i.e., each point is connected to two red lines and two blue lines (if there is a point $A$ connected to at least three lines of the same color $A B, A C, A D$, then $\triangle B C D$ is a monochromatic triangle of the other color, a contradiction). Assume the five numbers on the red circle are $a, b, c, d, e$ in sequence. Then $a+b, b+c, c+d, d+e, e+a$ are rational numbers.
$$
\text { By }(a+b)-(b+c)+(c+d)-(d+e)
$$
$+(e+a)=2a$ leads to a contradiction. Thus, the monochromatic triangle must be a blue triangle.
In summary, the minimum value of $n$ is 5.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. From the set $\{1,2, \cdots, 2011\}$, randomly select 1005 different numbers such that their sum is 1021035. Then, there are at least some odd numbers among them.
|
6.5.
From 1 to 2011, taking 1005 even numbers, their sum is $\frac{1005(2+2010)}{2}=1011030$.
Replacing $2,4,6,8,10$ with 2003, 2005, $2007,2009,2011$, increases the sum by
$$
2001 \times 5=10005,
$$
making the total sum 1021035.
Thus, there are at least 5 odd numbers.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. (20 points) Inside a large sphere with a radius of 4, 24 cubes with edge lengths of 1 have been placed arbitrarily. Prove: At least 4 small spheres with a radius of $\frac{1}{2}$ can still be placed inside the large sphere, such that these small spheres and the cubes are all within the large sphere and do not overlap with each other.
|
11. To place a small ball with a radius of $\frac{1}{2}$ completely inside the large ball $D(O, 4)$, the distance from its center to the surface of the large ball should be no less than $\frac{1}{2}$, meaning the center of the small ball should be within the ball $D\left(0,4-\frac{1}{2}\right)$. The volume of this ball is
$$
\frac{4 \pi}{3}\left(4-\frac{1}{2}\right)^{3}=\frac{343 \pi}{6}.
$$
Additionally, the center of the small ball should be no less than $\frac{1}{2}$ away from any surface of the cube. Therefore, the volume controlled by each cube within the large ball is
$$
3 \pi\left(\frac{1}{2}\right)^{2}+\frac{4 \pi}{3}\left(\frac{1}{2}\right)^{3}+4=\frac{11 \pi}{12}+4,
$$
which means the center of the small ball should not be in the $\frac{1}{4}$ cylinder (12 of them) with the vertex of the cube as the central axis and $\frac{1}{2}$ as the radius, the $\frac{1}{8}$ sphere (8 of them) with the vertex as the center and $\frac{1}{2}$ as the radius, the rectangular prism (6 of them) with the face as the base and $\frac{1}{2}$ as the height, and within the cube itself.
Similarly, the volume controlled by each small ball within the large ball is $\frac{4 \pi}{3}$.
If $l$ small balls have already been placed, the sufficient condition for placing the $(l+1)$-th small ball is
$$
\frac{343 \pi}{6}-24\left(\frac{11 \pi}{12}+4\right)-l \cdot \frac{4 \pi}{3} \geqslant 0,
$$
which simplifies to $l \leqslant \frac{3}{4}\left(\frac{343}{6}-22\right)-96 \cdot \frac{3}{4 \pi}$
$$
=\frac{211}{8}-\frac{72}{\pi} \approx 3.457.
$$
Therefore, at least 4 small balls can be placed.
|
4
|
Geometry
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
For any positive integer $n(n \geqslant 2)$, try to find:
$$
\begin{array}{l}
\sum_{k=2}^{n}\left[\log _{\frac{3}{2}}\left(k^{3}+1\right)-\log _{\frac{3}{2}}\left(k^{3}-1\right)\right]+ \\
\frac{1}{2} \log _{\frac{3}{2}}\left[1+\frac{1}{n^{2}}+\frac{1}{(n+1)^{2}}\right]
\end{array}
$$
the value.
|
Notice
$$
\begin{array}{l}
\sum_{k=2}^{n}\left[\log _{\frac{3}{2}}\left(k^{3}+1\right)-\log _{\frac{3}{2}}\left(k^{3}-1\right)\right] \\
=\sum_{k=2}^{n} \log _{\frac{3}{2}} \frac{k^{3}+1}{k^{3}-1}=\log _{\frac{3}{2}} \prod_{k=2}^{n} \frac{k^{3}+1}{k^{3}-1} \\
=\log _{\frac{3}{2}} \prod_{k=2}^{n} \frac{(k+1)\left(k^{2}-k+1\right)}{(k-1)\left(k^{2}+k+1\right)} \\
=\log _{\frac{3}{2}}\left(\prod_{k=2}^{n} \frac{k+1}{k-1} \cdot \prod_{k=2}^{n} \frac{k^{2}-k+1}{k^{2}+k+1}\right), \\
\prod_{k=2}^{n} \frac{k+1}{k-1}=\frac{n(n+1)}{2}, \\
\prod_{k=2}^{n} \frac{k^{2}-k+1}{k^{2}+k+1}=\prod_{k=2}^{n} \frac{(k-1)^{2}+(k-1)+1}{k^{2}+k+1} \\
=\frac{3}{n^{2}+n+1} .
\end{array}
$$
Also, $1+\frac{1}{n^{2}}+\frac{1}{(n+1)^{2}}=\frac{n^{4}+2 n^{3}+3 n^{2}+2 n+1}{n^{2}(n+1)^{2}}$
$$
=\left[\frac{n^{2}+n+1}{n(n+1)}\right]^{2} \text {, }
$$
then $\frac{1}{2} \log _{\frac{3}{2}}\left[1+\frac{1}{n^{2}}+\frac{1}{(1+n)^{2}}\right]$
$$
=\log _{\frac{3}{2}} \frac{n^{2}+n+1}{n(n+1)} \text {. }
$$
Substituting equations (2) and (3) into equation (1) gives
$$
\begin{array}{l}
\sum_{k=2}^{n}\left[\log _{\frac{3}{2}}\left(k^{3}+1\right)-\log _{\frac{3}{2}}\left(k^{3}-1\right)\right] \\
=\log _{\frac{3}{2}}\left[\frac{n(n+1)}{2} \cdot \frac{3}{n^{2}+n+1}\right] .
\end{array}
$$
From equations (4) and (5), we get
$$
\begin{array}{l}
\sum_{k=2}^{n}\left[\log _{\frac{3}{2}}\left(k^{3}+1\right)-\log _{\frac{3}{2}}\left(k^{3}-1\right)\right]+ \\
\quad \frac{1}{2} \log _{\frac{3}{2}}\left[1+\frac{1}{n^{2}}+\frac{1}{(n+1)^{2}}\right] \\
=\log _{\frac{3}{2}}\left[\frac{n(n+1)}{2} \cdot \frac{3}{n^{2}+n+1}\right]+\log _{\frac{3}{2}} \frac{n^{2}+n+1}{n(n+1)} \\
=\log _{\frac{3}{2}} \frac{3}{2}=1 .
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. Given the quadratic function
$$
y=a x^{2}+b x+c \geqslant 0(a<b) \text {. }
$$
Then the minimum value of $M=\frac{a+2 b+4 c}{b-a}$ is $\qquad$
|
11. 8.
From the conditions, it is easy to see that $a>0, b^{2}-4 a c \leqslant 0$.
Notice that
$$
\begin{array}{l}
M=\frac{a+2 b+4 c}{b-a}=\frac{a^{2}+2 a b+4 a c}{a(b-a)} \\
\geqslant \frac{a^{2}+2 a b+b^{2}}{a(b-a)} .
\end{array}
$$
Let $t=\frac{b}{a}$. Then $t>1$. Thus,
$$
\begin{array}{l}
M \geqslant \frac{a^{2}+2 a b+b^{2}}{a(b-a)}=\frac{t^{2}+2 t+1}{t-1} \\
=(t-1)+\frac{4}{t-1}+4 \\
\geqslant 2 \sqrt{4}+4=8 .
\end{array}
$$
The equality holds if and only if $t=3, b^{2}=$ $4 a c$, i.e., $b=3 a, c=\frac{9}{4} a$.
Therefore, the minimum value of $M$ is 8.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
16. (15 points) As shown in Figure 2, it is known that the ellipse $C$ passes through the point $M(2,1)$, with the two foci at $(-\sqrt{6}, 0)$ and $(\sqrt{6}, 0)$. $O$ is the origin, and a line $l$ parallel to $OM$ intersects the ellipse $C$ at two different points $A$ and $B$.
(1) Find the maximum value of the area of $\triangle OAB$;
(2) Prove that the lines $MA$, $MB$, and the $x$-axis form an isosceles triangle.
|
16. (1) Let the equation of the ellipse $C$ be
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0) \text {. }
$$
From the problem, we have
$$
\left\{\begin{array} { l }
{ a ^ { 2 } - b ^ { 2 } = 6 , } \\
{ \frac { 4 } { a ^ { 2 } } + \frac { 1 } { b ^ { 2 } } = 1 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a^{2}=8, \\
b^{2}=2 .
\end{array}\right.\right.
$$
Therefore, the equation of the ellipse is
$$
\frac{x^{2}}{8}+\frac{y^{2}}{2}=1 \text {. }
$$
Since the line $l \parallel O M$, we can set
$$
l: y=\frac{1}{2} x+m \text {. }
$$
Substituting the above equation into (1) gives
$$
x^{2}+2 m x+2 m^{2}-4=0 \text {. }
$$
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$. Then
$$
x_{1}+x_{2}=-2 m, x_{1} x_{2}=2 m^{2}-4 \text {. }
$$
Since the line $l$ intersects the ellipse $C$ at two distinct points $A$ and $B$, we have
$$
\Delta=(2 m)^{2}-4\left(2 m^{2}-4\right)>0 \text {. }
$$
Thus, $m \in(-2,2)$, and $m \neq 0$.
Therefore, $S_{\triangle O A B}=\frac{1}{2}|m|\left|x_{1}-x_{2}\right|$
$$
\begin{array}{l}
=\frac{1}{2}|m| \sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}} \\
=|m| \sqrt{4-m^{2}}=\sqrt{m^{2}\left(4-m^{2}\right)}
\end{array}
$$
$\leqslant 4$.
The equality holds if and only if $m^{2}=4-m^{2}$, i.e., $m= \pm \sqrt{2}$.
Therefore, the maximum area of $\triangle O A B$ is 4.
(2) Let the slopes of the lines $M A$ and $M B$ be $k_{1}$ and $k_{2}$, respectively. Then
$$
k_{1}=\frac{y_{1}-1}{x_{1}-2}, k_{2}=\frac{y_{2}-1}{x_{2}-2} .
$$
We only need to prove: $k_{1}+k_{2}=0$.
In fact,
$$
\begin{array}{l}
k_{1}+k_{2}=\frac{\frac{1}{2} x_{1}+m-1}{x_{1}-2}+\frac{\frac{1}{2} x_{2}+m-1}{x_{2}-2} \\
=1+m\left(\frac{1}{x_{1}-2}+\frac{1}{x_{2}-2}\right) \\
=1+m \cdot \frac{\left(x_{1}+x_{2}\right)-4}{x_{1} x_{2}-2\left(x_{1}+x_{2}\right)+4} \\
=1+m \cdot \frac{-2 m-4}{2 m^{2}-4-2(-2 m)+4}=0 .
\end{array}
$$
Therefore, the lines $M A$ and $M B$ form an isosceles triangle with the $x$-axis.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
17. (15 points) Given the function
$$
f(x)=\frac{1}{2} m x^{2}-2 x+1+\ln (x+1)(m \geqslant 1) \text {. }
$$
(1) If the curve $C: y=f(x)$ has a tangent line $l$ at point $P(0,1)$ that intersects $C$ at only one point, find the value of $m$;
(2) Prove that the function $f(x)$ has a decreasing interval $[a, b]$, and find the range of the length $t=b-a$ of the decreasing interval.
|
17. (1) Note that the domain of the function $f(x)$ is $(-1,+\infty)$,
$$
f^{\prime}(x)=m x-2+\frac{1}{x+1}, f^{\prime}(0)=-1 .
$$
Therefore, the slope of the tangent line $l$ at the point of tangency $P(0,1)$ is -1.
Thus, the equation of the tangent line is $y=-x+1$.
Since the tangent line $l$ intersects the curve $C$ at only one point, the equation
$$
\frac{1}{2} m x^{2}-x+\ln (x+1)=0
$$
has exactly one real solution.
Clearly, $x=0$ is a solution to the equation.
Let $g(x)=\frac{1}{2} m x^{2}-x+\ln (x+1)$. Then
$$
g^{\prime}(x)=m x-1+\frac{1}{x+1}=\frac{m x\left[x-\left(\frac{1}{m}-1\right)\right]}{x+1} \text {. }
$$
When $m=1$, $g^{\prime}(x)=\frac{x^{2}}{x+1} \geqslant 0$ (with equality only when $x=0$), so $g(x)$ is monotonically increasing on $(-1,+\infty)$, meaning $x=0$ is the only real solution to the equation.
When $m>1$, from
$$
g^{\prime}(x)=\frac{m x\left[x-\left(\frac{1}{m}-1\right)\right]}{x+1}=0 \text {, }
$$
we get $x_{1}=0, x_{2}=\frac{1}{m}-1 \in(-1,0)$.
On the interval $\left(-1, x_{2}\right)$, $g^{\prime}(x)>0$, and on the interval $\left(x_{2}, 0\right)$, $g^{\prime}(x)g(0)=0$.
As $x \rightarrow-1$, $g(x) \rightarrow-\infty$, so $g(x)=0$ has another solution in $\left(-1, x_{2}\right)$, which is a contradiction.
In summary, we have $m=1$.
(2) Note that
$$
\begin{array}{l}
f^{\prime}(x)=\frac{m x^{2}+(m-2) x-1}{x+1}(x>-1) . \\
\text { Hence } f^{\prime}(x)0$, and the axis of symmetry is
$$
\begin{array}{l}
x=-\frac{1}{2}+\frac{1}{m}>-1, \\
h(-1)=m-(m-2)-1=1>0,
\end{array}
$$
Therefore, the equation $h(x)=0$ has two distinct real roots $x_{1} 、 x_{2}$ in $(-1,+\infty)$, meaning the solution set of equation (1) is $\left(x_{1} 、 x_{2}\right)$.
Thus, the interval of monotonic decrease for the function $f(x)$ is $\left[x_{1}, x_{2}\right]$.
Then $t=x_{2}-x_{1}=\sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}}$
$$
=\sqrt{\frac{\Delta}{m^{2}}}=\sqrt{1+\frac{4}{m^{2}}} \text {. }
$$
Since $m \geqslant 1$, we have $1<\sqrt{1+\frac{4}{m^{2}}} \leqslant \sqrt{5}$.
Therefore, the range of the length $t$ of the interval of monotonic decrease for the function $y=f(x)$ is $(1, \sqrt{5}]$.
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Let real numbers $s, t$ satisfy
$$
\begin{array}{l}
19 s^{2}+99 s+1=0, \\
t^{2}+99 t+19=0(s t \neq 1) . \\
\text { Find the value of } \frac{s t+4 s+1}{t} \text { . }
\end{array}
$$
(1999, National Junior High School Mathematics Competition)
|
【Analysis】Transform the first equation of the known equations, and combine it with the second equation to find that $\frac{1}{s} 、 t(s t \neq 1)$ are the two roots of the quadratic equation
$$
x^{2}+99 x+19=0
$$
Since $s \neq 0$, the first equation can be transformed into
$$
\left(\frac{1}{s}\right)^{2}+99\left(\frac{1}{s}\right)+19=0 \text {. }
$$
Also, $t^{2}+99 t+19=0$, and $s t \neq 1$, so $\frac{1}{s} 、 t$ are the two distinct real roots of the quadratic equation
$$
x^{2}+99 x+19=0
$$
Thus, $\frac{1}{s}+t=-99, \frac{1}{s} \cdot t=19$, which means
$$
s t+1=-99 s, t=19 s . \\
\text { Therefore, } \frac{s t+4 s+1}{t}=\frac{-99 s+4 s}{19 s}=-5 .
$$
|
-5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given real numbers $x, y, z$ satisfy
$$
\left\{\begin{array}{l}
x+y=z-1, \\
x y=z^{2}-7 z+14 .
\end{array}\right.
$$
Question: What is the maximum value of $x^{2}+y^{2}$? For what value of $z$ does $x^{2}+y^{2}$ achieve its maximum value?
|
Prompt: Example 6. From the problem, we know that $x$ and $y$ are the two real roots of the equation
$$
t^{2}-(z-1) t+z^{2}-7 z+14=0
$$
By the discriminant $\Delta \geqslant 0$, we get
$$
3 z^{2}-26 z+55 \leqslant 0 \text {. }
$$
Solving this, we get $\frac{11}{3} \leqslant z \leqslant 5$.
$$
\begin{array}{l}
\text { Also, } x^{2}+y^{2}=(x+y)^{2}-2 x y \\
=(z-1)^{2}-2\left(z^{2}-7 z+14\right) \\
=(z-6)^{2}+9,
\end{array}
$$
Therefore, when $z=5$, the maximum value of $x^{2}+y^{2}$ is 9.
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
the integer part.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
|
Let
then $2<\sqrt[3]{23}=a_{1}<3, a_{n+1}=\sqrt[3]{23+a_{n}}$.
By mathematical induction, it is easy to prove
$$
2<a_{n}<3\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
In fact,
the integer part is 2.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given real numbers $x, y$ satisfy
$$
x^{2}+3 y^{2}-12 y+12=0 \text {. }
$$
then the value of $y^{x}$ is $\qquad$
|
Hint: Treat the known equation as a quadratic equation in $y$ (the main variable)
$$
3 y^{2}-12 y+\left(12+x^{2}\right)=0 \text {. }
$$
From $\Delta=-12 x^{2} \geqslant 0$, and since $x^{2} \geqslant 0$, then $x=0$.
Thus, $y=2$. Therefore, $y^{x}=2^{0}=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given real numbers $a, b, c$ satisfy
$$
a=2 b+\sqrt{2}, a b+\frac{\sqrt{3}}{2} c^{2}+\frac{1}{4}=0 \text {. }
$$
Then $\frac{b c}{a}=$ $\qquad$ .
|
Hint: Eliminate $a$ from the two known equations, then treat $c$ as a constant, to obtain a quadratic equation in $b$
$$
2 b^{2}+\sqrt{2} b+\frac{\sqrt{3}}{2} c^{2}+\frac{1}{4}=0 .
$$
From the discriminant $\Delta \geqslant 0$, we get $c=0$.
Therefore, $\frac{b c}{a}=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Given
$\frac{1}{4}(b-c)^{2}=(a-b)(c-a)$, and $a \neq 0$.
Then $\frac{b+c}{a}=$ $\qquad$
|
From the given equation, we have
$$
(b-c)^{2}-4(a-b)(c-a)=0 \text {. }
$$
When $a \neq b$, by the discriminant of a quadratic equation, we know that the quadratic equation in $x$
$$
(a-b) x^{2}+(b-c) x+(c-a)=0
$$
has two equal real roots.
$$
\text { Also, }(a-b)+(b-c)+(c-a)=0 \text {, thus, }
$$
$x=1$ is a root of equation (1).
Therefore, both roots of equation (1) are $x=1$.
So, $1 \times 1=\frac{c-a}{a-b} \Rightarrow \frac{b+c}{a}=2$.
When $a=b$, we have
$$
(b-c)^{2}=0 \Rightarrow b=c \text {. }
$$
In this case, $\frac{b+c}{a}=\frac{b+b}{b}=2$.
In summary, $\frac{b+c}{a}=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, (15 points) Can 2010 be written as the sum of squares of $k$ distinct prime numbers? If so, try to find the maximum value of $k$; if not, please briefly explain the reason.
|
If 2010 can be written as the sum of squares of $k$ prime numbers, taking the sum of the squares of the smallest 10 distinct primes, then
$$
\begin{array}{l}
4+9+25+49+121+169+ \\
289+361+529+841 \\
=2397>2010 .
\end{array}
$$
Therefore, $k \leqslant 9$.
It is easy to see that there is only one even prime number 2, and the rest are odd. The square of an odd number is congruent to 1 modulo 8.
Since 2010 is congruent to 2 modulo 8, but the sum of the squares of nine different primes is congruent to 1 or 4 modulo 8, and the sum of the squares of eight different primes is congruent to 0 or 3 modulo 8, hence $k \leqslant 7$.
When $k=7$, after trial calculation, we get
$$
2^{2}+3^{2}+7^{2}+11^{2}+13^{2}+17^{2}+37^{2}=2010 \text {. }
$$
In summary, 2010 can be written as the sum of squares of $k$ distinct prime numbers, and the maximum value of $k$ is 7.
In fact, it can also be proven that $k \neq 1,2, \cdots, 6$.
Therefore, 2010 can only be expressed as the sum of squares of seven distinct prime numbers.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given points $M(0,2)$ and $N(-3,6)$, the distances from these points to line $l$ are $1$ and $4$, respectively. The number of lines $l$ that satisfy these conditions is . $\qquad$
|
2.3.
It is easy to get $M N=5$.
Then the circle $\odot M$ with radius 1 is externally tangent to the circle $\odot N$ with radius 4. Therefore, there are 3 lines $l$ that satisfy the condition.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=2, a_{n+1}=\frac{1+a_{n}}{1-a_{n}}\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
Let $T_{n}=a_{1} a_{2} \cdots a_{n}$. Then $T_{2010}=$ $\qquad$
|
5. -6 .
It is easy to get $a_{1}=2, a_{2}=-3, a_{3}=-\frac{1}{2}, a_{4}=\frac{1}{3}$, $a_{1} a_{2} a_{3} a_{4}=1$.
Also, $a_{5}=2=a_{1}$, by induction it is easy to know $a_{n+4}=a_{n}\left(n \in \mathbf{N}_{+}\right)$.
Therefore, $T_{2010}=T_{4 \times 502+2}=a_{1} a_{2}=-6$.
|
-6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $a=\frac{\sqrt{5}-1}{2}$. Then $\frac{a^{5}+a^{4}-2 a^{3}-a^{2}-a+2}{a^{3}-a}=$ $\qquad$
|
Given $a=\frac{-1+\sqrt{5}}{2}$, we can obtain the corresponding quadratic equation $a^{2}+a-1=0$.
Thus, the original expression is
$$
\begin{array}{l}
=\frac{\left(a^{2}+a-1\right)\left(a^{3}-a\right)-2(a-1)}{a\left(a^{2}-1\right)} \\
=\frac{-2}{a^{2}+a}=-2 .
\end{array}
$$
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given the sequence $\left\{x_{n}\right\}$ :
$$
1,3,3,3,5,5,5,5,5, \cdots
$$
formed by all positive odd numbers arranged from smallest to largest, and each odd number $k(k=1,3,5, \cdots)$ appears consecutively $k$ times. If the general term formula of this sequence is $x_{n}=a[\sqrt{b n+c}]+d$, then $a+b+c+d=$ $\qquad$ ( $[x]$ denotes the greatest integer not exceeding the real number $x$).
|
4.3.
For $k^{2}+1 \leqslant n \leqslant(k+1)^{2}$, $x_{n}=2 k+1, k=[\sqrt{n-1}]$,
therefore, $x_{n}=2[\sqrt{n-1}]+1$.
Thus, $(a, b, c, d)=(2,1,-1,1)$.
So $a+b+c+d=3$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. $\sum_{i=0}^{50} \sum_{j=0}^{50} \mathrm{C}_{50}^{i} \mathrm{C}_{50}^{j}$ modulo 31 is
$\qquad$ .
|
6.1.
Original expression $=\sum_{i=0}^{50} C_{50}^{i} \cdot \sum_{j=0}^{50} C_{50}^{j}=\left(2^{50}\right)^{2}=2^{100}$.
And $2^{5} \equiv 1(\bmod 31)$, so
Original expression $=2^{100} \equiv 1(\bmod 31)$.
Therefore, the remainder is 1.
|
1
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. In a certain kingdom, there are 32 knights, some of whom are servants to other knights. Each servant can have at most one master, and each master must be richer than any of his servants. If a knight has at least four servants, he is ennobled as a noble. If it is stipulated that a servant of $A$'s servant is not a servant of $A$, then the maximum possible number of nobles is $\qquad$
|
2.7.
According to the problem, the richest knight is not a servant of any other knight, so at most 31 knights can be servants of other knights.
Since each noble has at least four servants, there can be at most 7 nobles.
Number the 32 knights as $1,2, \cdots, 32$, with wealth decreasing as the number increases, then
$$
\begin{array}{l}
1(2,3,4,5) ; \quad 5(6,7,8,9) ; \\
9(10,11,12,13) ; 13(14,15,16,17) ; \\
17(18,19,20,21) ; 21(22,23,24,25) ; \\
25(26,27,28,29),
\end{array}
$$
Each knight outside the parentheses is a noble, and the knights inside the parentheses are servants.
This shows that 7 nobles are possible.
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (25 points) Given 15 quadratic equations $x^{2}-p_{i} x+q_{i}=0(i=1,2, \cdots, 15)$ with coefficients $p_{i} 、 q_{i}$ taking values from $1,2, \cdots, 30$, and these coefficients are all distinct. If an equation has a root greater than 20, it is called a "good equation." Find the maximum number of good equations.
|
Second, if there exists an equation $x^{2}-p_{i} x+q_{i}=0$ with two roots $x_{1}, x_{2}$, then from $x_{1}+x_{2}=p_{i}>0, x_{1} x_{2}=q_{i}>0$, we know that both roots are positive.
Next, if a certain equation has a root greater than 20, then $p_{i}=x_{1}+x_{2}>20$. However, among the numbers $1,2, \cdots, 30$, there are only 10 numbers greater than 20, so the number of good equations is at most 10.
Finally, consider
$$
x^{2}-(20+k) x+k=0(k=1,2, \cdots, 10) \text {, }
$$
these ten equations. Their larger root
$$
\begin{array}{l}
x=\frac{20+k+\sqrt{(20+k)^{2}-4 k}}{2} \\
=\frac{20+k+\sqrt{k^{2}+36 k+400}}{2} \\
>\frac{20+k+k+18}{2}=k+19 \geqslant 20 .
\end{array}
$$
Therefore, the maximum number of good equations is 10.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) As shown in Figure 2, given points $A$ and $B$ are two distinct points outside circle $\odot O$, point $P$ is on $\odot O$, and $PA$, $PB$ intersect $\odot O$ at points $D$ and $C$ respectively, different from point $P$, and $AD \cdot AP = BC \cdot BP$.
(1) Prove: $\triangle OAB$ is an isosceles triangle;
(2) Let $p$ be a prime number, and $m$ be a positive integer. If $AD \cdot AP = p(2p + 1)$, $OA = m - 1$, and the radius of $\odot O$ is 3, find the length of $OA$.
|
(1) Draw $A T$ tangent to $\odot O$ at point $T$, and connect $O T$.
Let the radius of $\odot O$ be $R$. Then
$$
A D \cdot A P=A T^{2}=O A^{2}-R^{2} \text {. }
$$
Similarly, $B C \cdot B P=O B^{2}-R^{2}$.
From the given, it is easy to see that $O A=O B$.
(2) From the proof in (1), we know
$$
p(2 p+1)=(m-1)^{2}-9 \text {, }
$$
which means $p(2 p+1)=(m-4)(m+2)>0$.
Since $p$ is a prime and $(p, 2 p+1)=1$, we have
$$
p \mid(m-4) \text { or } p \mid(m+2)
$$
exactly one of them holds.
If $p \mid(m-4)$, then $m-4=k p\left(k \in \mathbf{N}_{+}\right)$.
If $k=1$, then
$$
2 p+1=p+6 \Rightarrow p=5, m=9, O A=8 \text {. }
$$
If $k \geqslant 2$, then $m+2=k p+6>2 p+1$, a contradiction.
If $p \mid(m+2)$, then $m+2=k p\left(k \in \mathbf{N}_{+}\right)$.
If $k \leqslant 2$, then $m-4=k p-6<2 p+1$, a contradiction.
Thus, $k \geqslant 3$. In this case,
$$
\begin{array}{l}
2 p+1=k(k p-6) \geqslant 3(3 p-6) \\
\Rightarrow 7 p \leqslant 19 \Rightarrow p=2 .
\end{array}
$$
However, when $p=2$, $(m-4)(m+2)=10$ has no positive integer solutions, a contradiction.
In summary, $O A=8$.
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given the inequality
$$
3 x+4 \sqrt{x y} \leqslant a(x+y)
$$
holds for all positive numbers $x$ and $y$. Then the minimum value of the real number $a$ is . $\qquad$
|
-1.4 .
From the problem, we know that $a \geqslant\left(\frac{3 x+4 \sqrt{x y}}{x+y}\right)_{\text {max }}$.
$$
\text { Also, } \frac{3 x+4 \sqrt{x y}}{x+y} \leqslant \frac{3 x+(x+4 y)}{x+y}=4 \text {, }
$$
with equality holding if and only if $x=4 y>0$.
Therefore, the minimum value of $a$ is 4.
|
4
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. $[x]$ is the greatest integer not exceeding the real number $x$. It is known that the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=\frac{3}{2}, a_{n+1}=a_{n}^{2}-a_{n}+1\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
Then $m=\left[\sum_{k=1}^{2011} \frac{1}{a_{k}}\right]$ is . $\qquad$
|
4.1.
$$
\begin{array}{l}
\text { Given } a_{n+1}=a_{n}^{2}-a_{n}+1 \\
\Rightarrow a_{n+1}-1=a_{n}\left(a_{n}-1\right) \\
\Rightarrow \frac{1}{a_{n+1}-1}=\frac{1}{a_{n}-1}-\frac{1}{a_{n}} . \\
\text { Then } \frac{1}{a_{n}}=\frac{1}{a_{n}-1}-\frac{1}{a_{n+1}-1} . \\
\text { Therefore } m=\left[\frac{1}{a_{1}-1}-\frac{1}{a_{2012}-1}\right]=\left[2-\frac{1}{a_{2012}-1}\right]=1 .
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Let $f(x)$ be an odd function defined on $\mathbf{R}$, and
$$
f(x)=f(1-x) \text {. }
$$
Then $f(2010)=$ $\qquad$ .
|
$$
\begin{array}{l}
f(0)=0, \\
f(x+1)=f(-x)=-f(x) .
\end{array}
$$
From the conditions, we have $f(x+2)=f(x)$, which means $f(x)$ is a periodic function with a period of 2.
Therefore, $f(2010)=f(0)=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Given real numbers $x, y$ satisfy
$$
3|x+1|+2|y-1| \leqslant 6 \text {. }
$$
Then the maximum value of $2 x-3 y$ is $\qquad$ .
|
10.4.
The figure determined by $3|x+1|+2|y-1| \leqslant 6$ is the quadrilateral $ABCD$ and its interior, where,
$$
A(-1,4) 、 B(1,1) 、 C(-1,-2) 、 D(-3,1) \text {. }
$$
By the knowledge of linear programming, the maximum value of $2x-3y$ is 4.
The maximum value can be achieved when $x=-1, y=-2$.
|
4
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let the sequence $\left\{8 \times\left(-\frac{1}{3}\right)^{n-1}\right\}$ have the sum of its first $n$ terms as $S_{n}$. Then the smallest integer $n$ that satisfies the inequality
$$
\left|S_{n}-6\right|<\frac{1}{125}
$$
is
|
2. 7
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. For $0<x<1$, if the complex number
$$
z=\sqrt{x}+\mathrm{i} \sqrt{\sin x}
$$
corresponds to a point, then the number of such points inside the unit circle is $n=$
|
6. 1 .
From the point on the unit circle, we have
$$
x+\sin x=1(00(x \in(0,1))$, which means $\varphi(x)$ is a strictly increasing function.
Also, $\varphi(0)=-10$, so the equation $x+\sin x=1$ has exactly one real root in $(0,1)$.
Therefore, $n=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Let $\frac{m}{n}=1+\frac{1}{2}+\cdots+\frac{1}{2010}$.
Then $m=$ $\qquad$ $(\bmod 2011)$.
|
8. 0 .
Notice
$$
\begin{array}{l}
\frac{m}{n}=1+\frac{1}{2}+\cdots+\frac{1}{2010} \\
=\left(\frac{1}{1}+\frac{1}{2010}\right)+\left(\frac{1}{2}+\frac{1}{2009}\right)+\cdots+ \\
\left(\frac{1}{1005}+\frac{1}{1006}\right) \\
= \frac{2011}{1 \times 2010}+\frac{2011}{2 \times 2009}+\cdots+\frac{2011}{1005 \times 1006} \\
= 2011 \times \frac{q}{2010!},
\end{array}
$$
where $q$ is a positive integer.
Thus, $2011 n q=m \times 2010!$.
This shows that 2011 divides $m \times 2010!$, but 2011 is a prime number and does not divide 2010!, so 2011 must divide $m$, giving $m \equiv 0(\bmod 2011)$.
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Consider the matrix
$$
\left(a_{i j}\right)_{n \times n}\left(a_{i j} \in\{1,2,3\}\right) \text {. }
$$
If $a_{i j}$ is such that its row $i$ and column $j$ both contain at least three elements (including $a_{i j}$) that are equal to $a_{i j}$, then the element $a_{i j}$ is called "good". If the matrix $\left(a_{i j}\right)_{n \times n}$ contains at least one good element, find the minimum value of $n$.
|
The minimum value of $n$ is 7.
When $n=6$, the matrix
$\left(\begin{array}{llllll}1 & 1 & 2 & 2 & 3 & 3 \\ 1 & 1 & 3 & 3 & 2 & 2 \\ 2 & 2 & 1 & 1 & 3 & 3 \\ 2 & 2 & 3 & 3 & 1 & 1 \\ 3 & 3 & 2 & 2 & 1 & 1 \\ 3 & 3 & 1 & 1 & 2 & 2\end{array}\right)$
has no good elements, so $n \geqslant 7$.
Below, we use proof by contradiction to show that when $n=7$, the matrix $\left(a_{i j}\right)$ must have at least one good element.
Assume that no element in the matrix $\left(a_{i j}\right)$ is good.
Define: If a number $m$ appears at least three times in a row (or column), then we say that the number $m$ controls that row (or column).
It is easy to see that the number of rows (or columns) controlled by the number $m$ is at most 4.
First, we prove a lemma:
Lemma: If the number 1 controls the first three rows, then the 1s in these three rows must appear in at least five columns.
Proof: Since the number 1 appears at least 9 times in the first three rows, by the pigeonhole principle, it must appear in at least five columns. Now, we show that five columns are impossible.
If the number 1 appears in the first five columns, consider the submatrix $T$ formed by the first five columns and the last four rows of the original matrix. Then, $T$ contains at most one 1.
If there is a row in $T$ where 2 appears at least three times and another row where 3 appears at least three times, then there must be a column where 1, 2, and 3 each appear three times. Therefore, in this column, 1, 2, and 3 each appear at most twice, which is a contradiction.
Assume that there are three rows in $T$ where 2 appears at least three times, then in the first five columns, the numbers 1 and 2 each appear at most twice in each column. Thus, in the first five columns, the number 3 appears at least three times in each column, which is a contradiction.
Therefore, this scenario is impossible.
We prove in two steps:
(1) If the number $m$ controls at least 3 rows (or columns), then the number of elements in the matrix $\left(a_{i j}\right)$ that are equal to $m$ is at most 16.
We discuss two cases (assume $m=1$ without loss of generality).
(i) The number 1 controls four rows.
Assume these are the first four rows, then the number 1 appears at least 12 times in the first four rows. By the pigeonhole principle, it must appear in at least six columns.
(1) If the number 1 appears in seven columns, then the number of 1s is at most 14.
(2) If the number 1 appears in six columns, then the number of 1s is at most $6 \times 2 + 3 = 15$.
(ii) The number 1 controls three rows.
Assume these are the first three rows. By the lemma, the number 1 appears in at least six columns.
(1) If the number 1 appears in seven columns, then the number of 1s is at most 14.
(2) If the number 1 appears in six columns, then the number of 1s is at most $6 \times 2 + 4 = 16$.
(2) For $m=1,2,3$, the number of rows or columns controlled by $m$ is at least 3.
By the pigeonhole principle, assume 1 controls at least 3 rows.
We show that the number of rows or columns controlled by 2 and 3 is at least 3.
We discuss two cases:
(i) The number 1 controls four rows.
Assume these are the first four rows, then the number 1 appears at least 12 times in the first four rows. By the pigeonhole principle, it must appear in at least six columns.
(1) If the number 1 appears in seven columns, then 2 and 3 each control three columns and four columns, respectively.
(2) If the number 1 appears in the first six columns, consider the submatrix $T$ formed by the first six columns and the last three rows of the original matrix. $T$ contains only 2 and 3.
If the first row of $T$ contains 3 twos and 3 threes, then 2 and 3 each control three columns; if the first row of $T$ contains 4 twos, then 3 controls four columns, and 2 controls three rows (since in the last two rows of $T$, it is impossible for one row to contain more than 3 threes).
(ii) The number 1 controls three rows.
Assume these are the first three rows. By the lemma, the number 1 appears in at least six columns.
(1) If the number 1 appears in seven columns, then 2 and 3 each control three columns and four columns, respectively.
(2) If the number 1 appears in the first six columns, consider the submatrix $T$ formed by the first six columns and the last four rows of the original matrix. $T$ contains at most 3 ones, so there must be a row in $T$ that does not contain 1.
If this row contains 3 twos and 3 threes, then 2 and 3 each control three columns; if this row contains 4 twos, then 3 controls four columns.
Since 1 appears at most 3 times, 2 must control at least three rows in the last four rows.
By (1) and (2), in the matrix $\left(a_{i j}\right)$, the number of rows or columns controlled by the numbers $m=1,2,3$ is at least 3.
Thus, the numbers 1, 2, and 3 each appear at most 16 times in the matrix $\left(a_{i j}\right)$.
But when $n=7$, the matrix $\left(a_{i j}\right)$ has 49 elements, which is a contradiction.
Therefore, when $n=7$, the matrix $\left(a_{i j}\right)$ must have at least one good element.
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. B. Given that $\angle A, \angle B$ are two acute angles, and satisfy
$$
\begin{array}{l}
\sin ^{2} A+\cos ^{2} B=\frac{5}{4} t, \\
\cos ^{2} A+\sin ^{2} B=\frac{3}{4} t^{2} .
\end{array}
$$
Then the sum of all possible real values of $t$ is ().
(A) $-\frac{8}{3}$
(B) $-\frac{5}{3}$
(C) 1
(D) $\frac{11}{3}$
|
3. B. C.
Adding the two equations yields $3 t^{2}+5 t=8$.
Solving gives $t=1, t=-\frac{8}{3}$ (discard).
When $t=1$, $\angle A=45^{\circ}, \angle B=30^{\circ}$ satisfies the given equations.
Therefore, the sum of all possible values of the real number $t$ is 1.
|
1
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
5. A. Let $S=\frac{1}{1^{3}}+\frac{1}{2^{3}}+\cdots+\frac{1}{99^{3}}$. Then the integer part of $4 S$ is ( ).
(A) 4
(B) 5
(C) 6
(D) 7
|
5. A. A.
When $k=2,3, \cdots, 99$, we have
$$
\frac{1}{k^{3}}<\frac{1}{k\left(k^{2}-1\right)}=\frac{1}{2}\left[\frac{1}{(k-1) k}-\frac{1}{k(k+1)}\right] \text {, }
$$
Thus, $1<S=1+\sum_{k=2}^{29} \frac{1}{k^{3}}$
$$
<1+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{99 \times 100}\right)<\frac{5}{4} \text {. }
$$
Therefore, $4<4 S<5$.
Hence, the integer part of $4 S$ is 4.
|
4
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
8. A. As shown in Figure 1, points $A$ and $B$ lie on the line $y=x$. Parallel lines to the $y$-axis through $A$ and $B$ intersect the hyperbola $y=\frac{1}{x} (x>0)$ at points $C$ and $D$. If $BD=2AC$, then the value of $4OC^2-OD^2$ is $\qquad$
|
8. A. 6 .
Let points $C(a, b), D(c, d)$.
Then points $A(a, a), B(c, c)$.
Since points $C, D$ are on the hyperbola $y=\frac{1}{x}$, we have $a b=1, c d=1$.
Given $B D=2 A C$
$$
\begin{array}{l}
\Rightarrow|c-d|=2|a-b| \\
\Rightarrow c^{2}-2 c d+d^{2}=4\left(a^{2}-2 a b+b^{2}\right) \\
\Rightarrow 4\left(a^{2}+b^{2}\right)-\left(c^{2}+d^{2}\right)=8 a b-2 c d=6 \\
\Rightarrow 4 O C^{2}-O D^{2}=6 .
\end{array}
$$
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. B. Let the four-digit number
$\overline{a b c d}$ satisfy
$$
a^{3}+b^{3}+c^{3}+d^{3}+1=10 c+d .
$$
Then the number of such four-digit numbers is $\qquad$
|
10. B. 5 .
From $d^{3} \geqslant d$ we know $c^{3}+1 \leqslant 10 c \Rightarrow 1 \leqslant c \leqslant 3$.
If $c=3$, then $a^{3}+b^{3}+d^{3}=2+d$.
Thus, $d=1$ or 0. Therefore, $a=b=1$, which gives us $1131, 1130$ as solutions.
If $c=2$, then $a^{3}+b^{3}+d^{3}=11+d$. Thus, $d \leqslant 2$.
When $d=2$, $a^{3}+b^{3}=5$, no solution;
When $d=1$ or 0, $a^{3}+b^{3}=11$, no solution.
If $c=1$, then $a^{3}+b^{3}+d^{3}=8+d$. Thus, $d \leqslant 2$.
When $d=2$, $a=b=1$, which gives us $1112$ as a solution;
When $d=1$ or 0, $a=2, b=0$, which gives us $2011, 2010$ as solutions.
In summary, there are 5 four-digit numbers that satisfy the conditions.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Let $f(x)$ be a polynomial with integer coefficients, $f(0)=11$, and there exist $n$ distinct integers $x_{1}, x_{2}, \cdots, x_{n}$, such that
$$
f\left(x_{1}\right)=f\left(x_{2}\right)=\cdots=f\left(x_{n}\right)=2010 .
$$
Then the maximum value of $n$ is
|
7.3.
Let $g(x)=f(x)-2010$. Then $x_{1}, x_{2}, \cdots, x_{n}$ are all roots of $g(x)=0$.
Thus, $g(x)=\prod_{i=1}^{n}\left(x-x_{i}\right) \cdot q(x)$,
where $q(x)$ is a polynomial with integer coefficients. Therefore,
$$
\begin{array}{l}
g(0)=11-2010=-1999 \\
=\prod_{i=1}^{n}\left(-x_{i}\right) q(0) .
\end{array}
$$
Since 1999 is a prime number, -1999 can be at most the product of three different integers, i.e., $-1 \times 1 \times 1999$, hence $n \leqslant 3$.
When $n=3$,
$$
\begin{array}{l}
g(x)=(x+1)(x-1)(x+1999), \\
f(x)=(x+1)(x-1)(x+1999)+2010 .
\end{array}
$$
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. (14 points) As shown in Figure 2, the corridor is 3 m wide, the angle between the corridors is 120°, the ground is level, and the ends of the corridor are sufficiently long. Question: What is the maximum length of a horizontal rod (neglecting its thickness) that can pass through the corridor?
|
II. 9. As shown in Figure 4, draw any horizontal line through the inner vertex $P$ of the corridor corner, intersecting the outer sides of the corridor at points $A$ and $B$. The length of a wooden rod that can pass through the corridor in a horizontal position is less than or equal to $AB$.
Let $\angle B A Q=\alpha$. Then
$$
\begin{array}{l}
\angle A B Q=60^{\circ}-\alpha, \\
A B=A P+P B=\frac{3}{\sin \alpha}+\frac{3}{\sin \left(60^{\circ}-\alpha\right)} .
\end{array}
$$
As $\alpha$ varies, the minimum value of the above expression is the maximum length of the wooden rod that can pass through the corridor in a horizontal position.
By the arithmetic mean inequality and product-to-sum formulas, we get
$$
\begin{array}{l}
A B \geqslant 6 \sqrt{\frac{1}{\sin \alpha \cdot \sin \left(60^{\circ}-\alpha\right)}} \\
=6 \sqrt{\frac{1}{\frac{1}{2}\left[\cos \left(2 \alpha-60^{\circ}\right)-\cos 60^{\circ}\right]}} \\
\geqslant 6 \sqrt{\frac{2}{1-\frac{1}{2}}}=12,
\end{array}
$$
and when $\alpha=30^{\circ}$, $A B=12$.
Therefore, the maximum length of a wooden rod that can pass through the corridor in a horizontal position is $12 \mathrm{~m}$.
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given $a>0$, the graphs of the functions $f(x)=|x+2a|$ and $g(x)=|x-a|$ intersect at point $C$, and they intersect the $y$-axis at points $A$ and $B$ respectively. If the area of $\triangle ABC$ is 1, then $a=$ $\qquad$ .
|
2. 2 .
From the graphs of $f(x)$ and $g(x)$, we know that $\triangle ABC$ is an isosceles right triangle with base $a$, so its area is $\frac{a^{2}}{4}=1$. Therefore, $a=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given the function $y=x^{3}$, the tangent line at $x=a_{k}$ intersects the $x$-axis at point $a_{k+1}$. If $a_{1}=1, S_{n}=\sum_{i=1}^{n} a_{i}$, then $\lim _{n \rightarrow \infty} S_{n}$ $=$ . $\qquad$
|
4.3.
It is known that $y^{\prime}=3 x^{2}$.
Therefore, the equation of the tangent line to $y=x^{3}$ at $x=a_{k}$ is $y-a_{k}^{3}=3 a_{k}^{2}\left(x-a_{k}\right)$.
Thus, the above equation intersects the $x$-axis at the point $\left(a_{k+1}, 0\right)$. Hence, $-a_{k}^{3}=3 a_{k}^{2}\left(a_{k+1}-a_{k}\right)$.
From this, we get $a_{k+1}=\frac{2}{3} a_{k}$.
Given $a_{1}=1$, thus $\lim _{n \rightarrow \infty} S_{n}=\frac{1}{1-\frac{2}{3}}=3$.
|
3
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. The function $f: \mathbf{R} \rightarrow \mathbf{R}$ satisfies for all $x, y, z \in \mathbf{R}$
$$
f(x+y)+f(y+z)+f(z+x) \geqslant 3 f(x+2 y+z) .
$$
Then $f(1)-f(0)=$ $\qquad$
|
5.0.
Let $x=-y=z$, we get
$$
f(2 x) \geqslant f(0) \Rightarrow f(1) \geqslant f(0) \text {. }
$$
Let $x=y=-z$, we get
$$
f(0) \geqslant f(2 x) \Rightarrow f(0) \geqslant f(1) \text {. }
$$
Thus, $f(1)-f(0)=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. (18 points) Let $S$ be a set of distinct quadruples $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)$, where $a_{i}=0$ or 1 $(i=1,2,3,4)$. It is known that the number of elements in $S$ does not exceed 15, and satisfies:
if $\left(a_{1}, a_{2}, a_{3}, a_{4}\right) 、\left(b_{1}, b_{2}, b_{3}, b_{4}\right) \in S$, then $\left(c_{1}, c_{2}, c_{3}, c_{4}\right) 、\left(d_{1}, d_{2}, d_{3}, d_{4}\right) \in S$,
where, $c_{i}=\max \left\{a_{i}, b_{i}\right\}, d_{i}=\min \left\{a_{i}, b_{i}\right\}(i=1,2,3,4)$. Find the maximum number of elements in the set $S$.
|
12. Clearly, there are 16 possible quadruples. Since at least one quadruple is not in $S$, it follows that $(1,0,0,0)$, $(0,1,0,0)$, $(0,0,1,0)$, $(0,0,0,1)$ must have at least one that is not in $S$. Otherwise, by the given conditions, all quadruples would be in $S$.
Assume $(1,0,0,0) \notin S$.
In this case, by the given conditions,
$(1,1,0,0)$, $(1,0,1,0)$, $(1,0,0,1)$
at least two of these cannot be in $S$ (assume $(1,1,0,0)$ and $(1,0,1,0)$ are not in $S$). Then $(1,1,1,0)$ and $(1,1,0,1)$ cannot both be in $S$ (assume $(1,1,1,0)$ is not in $S$). Therefore, the number of elements in $S$ is at most $16-4=12$.
Let $S$ be the set of all 16 possible quadruples after removing the above 4 quadruples.
Next, we prove by contradiction that $S$ satisfies the conditions of the problem.
Take any $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)$, $\left(b_{1}, b_{2}, b_{3}, b_{4}\right) \in S$.
(1) If $\left(c_{1}, c_{2}, c_{3}, c_{4}\right) \notin S$, then $c_{1}=1$, $c_{4}=0$.
Thus, $1 \in \{a_{1}, b_{1}\}$, $a_{4}=b_{4}=0$.
Assume $a_{1}=1$, then $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)$ is among the 4 quadruples removed, which is a contradiction.
(2) If $\left(d_{1}, d_{2}, d_{3}, d_{4}\right) \notin S$, then $d_{1}=1$, $d_{4}=0$.
Thus, $a_{1}=b_{1}=1$, $0 \in \{a_{4}, b_{4}\}$.
Assume $a_{4}=0$, then $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)$ is among the 4 quadruples removed, which is a contradiction.
|
12
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Find the units digit of $\left(7^{2004}+36\right)^{818}$.
$(2004$, Shanghai Jiao Tong University Independent Recruitment Examination)
|
$$
\begin{array}{l}
\text { Solution: Since } 7^{4} \equiv 1(\bmod 10) \\
\Rightarrow 7^{2004} \equiv 1(\bmod 10) \\
\Rightarrow 7^{2004}+36 \equiv 7(\bmod 10) .
\end{array}
$$
$$
\text { Therefore, the original expression } \equiv 7^{818} \equiv 7^{2} \equiv 9(\bmod 10) \text {. }
$$
Thus, the unit digit is 9.
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Let $u$ be a root of the equation
$$
x^{3}-3 x+10=0
$$
Let $f(x)$ be a quadratic polynomial with rational coefficients, and
$$
\alpha=\frac{1}{2}\left(u^{2}+u-2\right), f(\alpha)=u .
$$
Find $f(0)$.
(2010, Five Schools Joint Examination for Independent Enrollment)
|
From the given, we have $u^{3}-3 u+10=0$. Then
$$
\begin{aligned}
\alpha^{2} &=\frac{1}{4}\left(u^{2}+u-2\right)^{2} \\
&=\frac{1}{4}\left(u^{4}+2 u^{3}-3 u^{2}-4 u+4\right) \\
&=\frac{1}{4}\left[u(3 u-10)+2(3 u-10)-3 u^{2}-4 u+4\right] \\
&=-4-2 u .
\end{aligned}
$$
Let $f(x)=a x^{2}+b x+c(a, b, c \in \mathbf{Q}, a \neq 0)$. Then
$$
\begin{aligned}
u &=f(\alpha)=a \alpha^{2}+b \alpha+c \\
&=a(-4-2 u)+\frac{b}{2}\left(u^{2}+u-2\right)+c \\
&\Rightarrow b u^{2}+(b-4 a-2) u-(8 a+2 b-2 c)=0 . \\
\text { Let } k &=(b-4 a-2), \\
l & =-(8 a+2 b-2 c) .
\end{aligned}
$$
$$
\begin{aligned}
\text { Then } b u^{2}+k u+l &=0(k, l \in \mathbf{Q}) . \\
\text { Hence } (b u-k)\left(b u^{2}+k u+l\right) &=
\end{aligned}
$$
$$
\begin{aligned}
&=b^{2} u^{3}+\left(b l-k^{2}\right) u-k l \\
&=b^{2}(3 u-10)+\left(b l-k^{2}\right) u-k l \\
&=\left(3 b^{2}+b l-k^{2}\right) u-\left(10 b^{2}+k l\right)=0 .
\end{aligned}
$$
Since $\pm 1, \pm 2, \pm 5, \pm 10$ are not roots of equation (1), then $u$ cannot be a rational number. Therefore,
$$
\left\{\begin{array}{l}
3 b^{2}+b l-k^{2}=0 \\
10 b^{2}+k l=0 .
\end{array}\right.
$$
Assume $b \neq 0$. Then,
$$
\begin{aligned}
k^{3} &=3 b^{2} k+b k l=3 b^{2} k-10 b^{3} \\
&\Rightarrow\left(\frac{k}{b}\right)^{3}-3\left(\frac{k}{b}\right)+10=0 .
\end{aligned}
$$
Thus, $\frac{k}{b}$ is a rational root of equation (1), which is impossible, so $b=0$.
$$
\begin{aligned}
\text { By } b u^{2}+k u+l &=0 \Rightarrow k u+l=0 \\
&\Rightarrow\left\{\begin{array}{l}
k=0, \\
l=0
\end{array} \Rightarrow \left\{\begin{array}{l}
-4 a-2=0, \\
8 a-2 c=0
\end{array}\right.\right. \\
&\Rightarrow\left\{\begin{array}{l}
a=-\frac{1}{2}, \\
c=-2 .
\end{array}\right.
\end{aligned}
$$
Therefore, $f(x)=-\frac{1}{2} x^{2}-2, f(0)=-2$.
In advanced algebra, a polynomial that cannot be expressed as the product of polynomials with rational coefficients is called an irreducible polynomial over the field of rational numbers, such as $x^{3}-3 x+10$ in equation (1), which is an irreducible polynomial over the field of rational numbers. By the degree theory in advanced algebra, we know that $x^{3}-3 x+10$ is the minimal polynomial of $u$. Hence, $u^{2}, u, 1$ are linearly independent over the field of rational numbers. Therefore, from $b u^{2}+k u+l=0$, we can deduce that $b=k=l=0$, which leads to the conclusion.
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Given real numbers $x, y, z$ satisfy
$$
x y z=32, x+y+z=4 \text {. }
$$
Then the minimum value of $|x|+|y|+|z|$ is $\qquad$ .
(2010, Hubei Province High School Mathematics Competition)
|
Answer: 12.
The text above has been translated into English, maintaining the original text's line breaks and format. Here is the direct output of the translation result.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given positive integers $a, b$ satisfy
$$
|b-2|+b-2=0,|a-b|+a-b=0,
$$
and $a \neq b$. Then the value of $a b$ is $\qquad$ .
|
2.1.2.
From the given conditions, we know that $a>0, b-2 \leqslant 0, a-b \leqslant 0$. Therefore, $a<b \leqslant 2$. Hence, $a=1, b=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. As shown in Figure 3, in $\triangle ABC$, it is given that $D$ is a point on side $BC$ such that $AD = AC$, and $E$ is the midpoint of side $AD$ such that $\angle BAD = \angle ACE$. If $S_{\triangle BDE} = 1$, then $S_{\triangle ABC}$ is $\qquad$.
|
4.4.
Notice that
$$
\begin{array}{l}
\angle E C D=\angle A C D-\angle A C E \\
=\angle A D C-\angle B A D=\angle A B C .
\end{array}
$$
Therefore, $\triangle E C D \backsim \triangle A B C$.
$$
\text { Then } \frac{C D}{B C}=\frac{E D}{A C}=\frac{E D}{A D}=\frac{1}{2} \text {. }
$$
Thus, $D$ is the midpoint of side $B C$.
Hence $S_{\triangle A B C}=2 S_{\triangle A C D}=4 S_{\triangle E C D}=4 S_{\triangle D B E}=4$.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (20 points) For a certain project, Team A alone needs 12 days to complete, and Team B alone needs 9 days to complete. If the two teams are scheduled to work in whole days, how many schemes are there to ensure that the project is completed within 8 days?
|
Three, let teams A and B work for $x, y$ days respectively to just meet the requirements. Then we have
$$
\left\{\begin{array}{l}
\frac{x}{12}+\frac{y}{9}=1, \\
0 \leqslant x \leqslant 8, \\
0 \leqslant y \leqslant 8,
\end{array}\right.
$$
and $x, y$ are both integers.
From $\frac{x}{12}+\frac{y}{9}=1$, we get $x=12-y-\frac{y}{3}$.
Since $x$ and $y$ are both integers, and $0 \leqslant y \leqslant 8$, so, $y=0$ or 3 or 6.
When $y=0$, $x=12$ (discard);
When $y=3$, $x=8$;
When $y=6$, $x=4$.
Therefore, there are two solutions, one is A works for 8 days, B works for 3 days; the other is A works for 4 days, B works for 6 days.
|
2
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let the polynomial $f(x)$ satisfy: for any $x \in \mathbf{R}$, we have
$$
f(x+1)+f(x-1)=2 x^{2}-4 x .
$$
Then the minimum value of $f(x)$ is $\qquad$
|
- 1. - -2 .
Since $f(x)$ is a polynomial, we know that $f(x+1)$ and $f(x-1)$ have the same degree as $f(x)$. Therefore, $f(x)$ is a quadratic polynomial (let's assume $f(x) = ax^2 + bx + c$). Then,
$$
\begin{array}{l}
f(x+1) + f(x-1) \\
= 2ax^2 + 2bx + 2(a + c) = 2x^2 - 4x.
\end{array}
$$
Thus, $a = 1, b = -2, c = -1$.
Hence, $f(x) = x^2 - 2x - 1 = (x-1)^2 - 2 \geqslant -2$.
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. If real numbers $x, y$ satisfy the system of equations
$$
\left\{\begin{array}{l}
(x-1)^{2011}+(x-1)^{2009}+2010 x=4020, \\
(y-1)^{2011}+(y-1)^{2009}+2010 y=0,
\end{array}\right.
$$
then $x+y=$ $\qquad$ .
|
2. 2 .
Transform the original system of equations into
$$
\begin{array}{l}
\left\{\begin{array}{l}
(x-1)^{2011}+(x-1)^{2009}+2010(x-1)=2010, \\
(y-1)^{2011}+(y-1)^{2009}+2010(y-1)=-2010 .
\end{array}\right. \\
\text { Let } f(t)=t^{2011}+t^{2009}+2010 t(t \in \mathbf{R}) .
\end{array}
$$
Then the function $f(t)$ is a monotonically increasing odd function.
From $f(x-1)=-f(y-1)$, we get
$$
x-1=-(y-1) \Rightarrow x+y=2 .
$$
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given $a, b, c \in \mathbf{R}$, and
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c} \text {, }
$$
then there exists an integer $k$, such that the following equations hold for
$\qquad$ number of them.
(1) $\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{2 k+1}=\frac{1}{a^{2 k+1}}+\frac{1}{b^{2 k+1}}+\frac{1}{c^{2 k+1}}$;
(2) $\frac{1}{a^{2 k+1}}+\frac{1}{b^{2 k+1}}+\frac{1}{c^{2 k+1}}=\frac{1}{a^{2 k+1}+b^{2 k+1}+c^{2 k+1}}$;
(3) $\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{2 k}=\frac{1}{a^{2 k}}+\frac{1}{b^{2 k}}+\frac{1}{c^{2 k}}$;
(4) $\frac{1}{a^{2 k}}+\frac{1}{b^{2 k}}+\frac{1}{c^{2 k}}=\frac{1}{a^{2 k}+b^{2 k}+c^{2 k}}$.
|
3. 2 .
From the given equation, we have
$$
\begin{array}{l}
\frac{(b c+a c+a b)(a+b+c)-a b c}{a b c(a+b+c)}=0 . \\
\text { Let } P(a, b, c) \\
=(b c+a c+a b)(a+b+c)-a b c \\
=(a+b)(b+c)(c+a) .
\end{array}
$$
From $P(a, b, c)=0$, we get
$$
a=-b \text { or } b=-c \text { or } c=-a \text {. }
$$
Verification shows that equations (1) and (2) hold, while equations (3) and (4) do not hold.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given $k_{1}, k_{2}, \cdots, k_{n}$ are $n$ distinct positive integers, and satisfy $\sum_{i=1}^{n} k_{i}^{3}=2024$. Then the maximum value of the positive integer $n$ is $\qquad$ .
|
6. 8 .
From the problem, we know that when $n \geqslant 9$, we have
$$
\sum_{i=1}^{n} k_{i}^{3} \geqslant \sum_{i=1}^{n} i^{3} \geqslant \sum_{i=1}^{9} i^{3}=2025>2024,
$$
which is a contradiction. Therefore, $n \leqslant 8$.
Also, $2^{3}+3^{3}+\cdots+9^{3}=2024$, so the maximum value of the positive integer $n$ is 8.
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Let $m, n$ be given positive integers. In each square of an $m \times n$ chessboard, fill a number according to the following rule: first, fill the numbers in the 1st row and the 1st column arbitrarily, then, for any other square, let the number filled in it be $x$, and the number in the 1st row in the same column as $x$ be $y$, the number in the 1st column in the same row as $x$ be $z$, and the number in the 1st row and 1st column be $a$, then $a + x = y + z$. Thus, a unique number can be filled in each square. After all squares are filled with numbers, take any rectangle on the chessboard, then the probability that the sum of the numbers in the two diagonal corners of the rectangle is equal is $\qquad$
|
6.1.
Let the cell at the $i$-th row and $j$-th column of the chessboard be denoted as $a_{i j}$, and the number filled in cell $a_{i j}$ is also represented by $a_{i j}$.
Consider any rectangle on the chessboard, and let the four corner cells of this rectangle be $a_{i j}, a_{i t}, a_{s j}, a_{s t} (i<s, j<t)$.
By the rule of filling numbers, we have
$$
\begin{array}{l}
a_{i j}=a_{1 j}+a_{i 1}-a_{11}, a_{i t}=a_{1 t}+a_{i 1}-a_{11}, \\
a_{s j}=a_{1 j}+a_{s 1}-a_{11}, a_{s t}=a_{1 t}+a_{s 1}-a_{11} . \\
\text { Then } a_{i j}+a_{s t}=\left(a_{1 j}+a_{i 1}-a_{11}\right)+\left(a_{1 t}+a_{s 1}-a_{11}\right) \\
=a_{1 j}+a_{i 1}+a_{1 t}+a_{s 1}-2 a_{11}, \\
a_{i t}+a_{s j}=\left(a_{1 t}+a_{i 1}-a_{11}\right)+\left(a_{1 j}+a_{s 1}-a_{11}\right) \\
=a_{1 j}+a_{i 1}+a_{1 t}+a_{s 1}-2 a_{11} .
\end{array}
$$
Therefore, $a_{i j}+a_{s t}=a_{i t}+a_{s j}$
Thus, the required probability is 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given $p$ is a prime number greater than 5, and $m$ is the smallest non-negative remainder when $\left(p^{2}+5 p+5\right)^{2}$ is divided by 120. Then the units digit of $2009^{m}$ is $\qquad$
|
Notice that
$$
\begin{array}{l}
\left(p^{2}+5 p+5\right)^{2}=\left[\left(p^{2}+5 p+5\right)^{2}-1\right]+1 \\
=\left(p^{2}+5 p+6\right)\left(p^{2}+5 p+4\right)+1 \\
=(p+1)(p+2)(p+3)(p+4)+1 .
\end{array}
$$
Let $M=(p+1)(p+2)(p+3)(p+4)$.
Clearly, $5! \mid p M$.
Since $p$ is a prime number greater than 5, 120 divides $M$.
Therefore, $\left(p^{2}+5 p+5\right)^{2}$ always has a remainder of 1 when divided by 120, i.e., $m=1$.
Thus, $2009^{m}=2009$.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let $D$, $E$, and $F$ be points on the sides $BC$, $AB$, and $AC$ of $\triangle ABC$, respectively, such that $AE = AF$, $BE = BD$, and $CF = CD$. Given that $AB \cdot AC = 2BD \cdot DC$, $AB = 12$, and $AC = 5$. Then the inradius $r$ of $\triangle ABC$ is $\qquad$
|
4.2.
As shown in Figure 6, let
$$
\begin{array}{l}
A E=A F=x, \\
B E=B D=y, \\
C D=C F=z .
\end{array}
$$
Then $A B=x+y$,
$$
\begin{array}{l}
A C=x+z, \\
B C=y+z .
\end{array}
$$
From $A B \cdot A C=2 B D \cdot D C$
$$
\begin{array}{l}
\Rightarrow(x+y)(x+z)=2 y z \\
\Rightarrow x^{2}+x y+x z=y z \\
\text { Also } A B^{2}+A C^{2}=(x+y)^{2}+(x+z)^{2} \\
=y^{2}+z^{2}+2\left(x^{2}+x y+x z\right) \\
=y^{2}+z^{2}+2 y z \\
=(y+z)^{2}=B C^{2},
\end{array}
$$
Then $\angle A=90^{\circ}$, and
$$
B C=\sqrt{A B^{2}+A C^{2}}=13 \text {. }
$$
Thus $r=\frac{A B \cdot A C}{A B+B C+C A}=2$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given positive real numbers $a$, $b$, $c$ satisfy
$$
(1+a)(1+b)(1+c)=8 \text {. }
$$
Then the minimum value of $a b c+\frac{9}{a b c}$ is
|
-1.10 .
$$
\begin{array}{l}
\text { Given } 8=(1+a)(1+b)(1+c) \\
\geqslant 2 \sqrt{a} \cdot 2 \sqrt{b} \cdot 2 \sqrt{c}=8 \sqrt{a b c} \\
\Rightarrow a b c \leqslant 1 .
\end{array}
$$
Equality holds if and only if $a=b=c=1$.
It is easy to verify that the function
$$
f(x)=x+\frac{9}{x}
$$
is decreasing on $(0,3)$.
Since $0<a b c \leqslant 1<3$, when $a b c=1$,
$$
f(a b c)=a b c+\frac{9}{a b c}
$$
is minimized, and the minimum value is 10.
|
10
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given a finite arithmetic sequence $\left\{a_{n}\right\}$ with the first term $a_{1}=1$, and a common difference of 2, the arithmetic mean of all its terms is 2011. If one term is removed, the arithmetic mean of the remaining terms is an integer. Then the number of ways to remove a term is $\qquad$.
|
4.3.
According to the problem, we have
$$
\frac{1}{n}\left[n+\frac{n(n-1)}{2} \times 2\right]=2011 \text {. }
$$
Solving this, we get $n=2011$.
Thus, the sum of all terms in the sequence is $2011^{2}$.
Suppose that after removing the $k$-th term from the sequence, the arithmetic mean of the remaining terms is an integer, i.e.,
$$
\frac{2011^{2}-(2 k-1)}{2010} \in \mathbf{Z} \text {. }
$$
Noting that $2011^{2}=2010 \times 2012+1$, the above equation is equivalent to
$$
2010|2(1-k) \Rightarrow 1005|(1-k) \text {. }
$$
Since $1 \leqslant k \leqslant 2011$, the set of possible values for $k$ is $\{1,1006,2011\}$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. As shown in Figure 2, in plane $\alpha$, $\triangle A B C$ and $\triangle A_{1} B_{1} C_{1}$ are on opposite sides of line $l$, with no common points with $l$, and are symmetric about line $l$. Now, if plane $\alpha$ is folded along line $l$ to form a right dihedral angle, then the six points $A, B, C, A_{1}, B_{1}, C_{1}$ can determine $\qquad$ planes (answer with a number).
|
7.11.
Notice that after the fold, the three sets of four points
$$
\left(A, B, A_{1}, B_{1}\right) 、\left(B, C, B_{1}, C_{1}\right) 、\left(C, A, C_{1}, A_{1}\right)
$$
are all coplanar, therefore, these six points can determine
$$
C_{6}^{3}-3\left(C_{4}^{3}-1\right)=11 \text { (planes). }
$$
|
11
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given points $A, B, C, D$ lie on the same circle, and $BC = DC = 4, AC$ intersects $BD$ at point $E, AE = 6$. If the lengths of segments $BE$ and $DE$ are both integers, find the length of $BD$.
|
Apply the property to $\triangle B C D$ to get
$$
C E^{2}=C D^{2}-B E \cdot E D=16-6 \cdot E C \text {. }
$$
Solving, we get $E C=2$.
From $B E \cdot E D=A E \cdot E C=12$, and
$$
B D<B C+C D=8 \text {, }
$$
we solve to get $B D=4+3=7$.
|
7
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. A circle is filled with 12 positive integers, each taken from $\{1,2, \cdots, 9\}$ (each number can appear multiple times on the circle). Let $S$ represent the sum of all 12 numbers on the circle. If the sum of any three consecutive numbers on the circle is a multiple of 7, then the number of possible values for $S$ is $\qquad$ kinds.
|
4.9.
For any three consecutive numbers $a_{k}, a_{k+1}, a_{k+2}$ on a circle, $a_{k}+a_{k+1}+a_{k+2}$ can be 7, 14, or 21.
For any four consecutive numbers on the circle, if they are $a_{k}, a_{k+1}, a_{k+2}, a_{k+3}$, since
$$
a_{k}+a_{k+1}+a_{k+2} \text { and } a_{k+1}+a_{k+2}+a_{k+3}
$$
are both multiples of 7, it must be that
$7 \mid\left(a_{k+3}-a_{k}\right)$.
Thus, $a_{k}$ and $a_{k+3}$ are either equal or differ by 7.
Now, divide the circle into four segments, each with a sum of three consecutive numbers that can be 7, 14, or 21, so the total sum of the four segments can be any of the following:
$$
\{28,35,42,49,56,63,70,77,84\}
$$
There are a total of 9 possible cases.
One way to fill the circle is: first, sequentially fill 12 numbers on the circle:
$$
1,2,4,1,2,4,1,2,4,1,2,4,
$$
with a sum of 28; then, each time change one 1 to 8, or one 2 to 9, each operation increases the total sum by 7. Such operations can be performed 8 times.
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. (20 points) Let
$$
P=x^{4}+6 x^{3}+11 x^{2}+3 x+31 \text {. }
$$
Find the integer value(s) of $x$ that make $P$ a perfect square.
|
10. Since $P=\left(x^{2}+3 x+1\right)^{2}-3(x-10)$, therefore, when $x=10$, $P=131^{2}$ is a perfect square.
Next, we only need to prove: there are no other integer $x$ that satisfy the requirement.
(1) $x>10$.
If $P0$, then $P>\left(x^{2}+3 x\right)^{2}$.
Therefore, $\left(x^{2}+3 x\right)^{2}\left(x^{2}+3 x+1\right)^{2}$.
Let $P=y^{2}(y \in \mathbf{Z})$. Then
$$
|y|>\left|x^{2}+3 x+1\right| \text {, }
$$
which means $|y|-1 \geqslant \mid x^{2}+3 x+11$.
Thus, $y^{2}-2|y|+1 \geqslant\left(x^{2}+3 x+1\right)^{2}$, i.e., $-3(x-10)-2\left|x^{2}+3 x+1\right|+1 \geqslant 0$.
Solving this inequality yields the integer values of $x$ as
$$
\pm 2, \pm 1, 0, -3, -4, -5, -6 \text {. }
$$
However, the corresponding $P$ for these values are not perfect squares.
In conclusion, the integer value of $x$ that makes $P$ a perfect square is 10.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. In the sequence $\left\{a_{n}\right\}$, it is known that
$$
a_{1}=2, a_{n+1}-2 a_{n}=2^{n+1}\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
Then the smallest positive integer $n$ for which $a_{n}>10$ holds is $\qquad$ . .
|
3.3.
Given $a_{n+1}-2 a_{n}=2^{n+1}$, we know $\frac{a_{n+1}}{2^{n+1}}-\frac{a_{n}}{2^{n}}=1$.
Thus, the sequence $\left\{\frac{a_{n}}{2^{n}}\right\}$ is an arithmetic sequence with a common difference of 1. Also, $a_{1}=2$, so $\frac{a_{n}}{2^{n}}=n \Rightarrow a_{n}=n \times 2^{n}$. Therefore, $\left\{a_{n}\right\}$ is an increasing sequence, and
$$
a_{2}=2 \times 2^{2}=8, a_{3}=3 \times 2^{3}=24 \text {. }
$$
Hence, the smallest positive integer $n$ is 3.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. If the positive integer $m$ makes it true that for any set of positive numbers $a_{1} 、 a_{2} 、 a_{3} 、 a_{4}$ satisfying $a_{1} a_{2} a_{3} a_{4}=1$, we have
$$
a_{1}^{m}+a_{2}^{m}+a_{3}^{m}+a_{4}^{m} \geqslant \frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\frac{1}{a_{4}}
$$
then the minimum value of the positive integer $m$ is $\qquad$
|
9.3.
Let $a_{1}=\frac{1}{27}, a_{2}=a_{3}=a_{4}=3$. Then
$$
\begin{array}{l}
a_{1}^{m}+a_{2}^{m}+a_{3}^{m}+a_{4}^{m}=\left(\frac{1}{27}\right)^{m}+3 \times 3^{m}, \\
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\frac{1}{a_{4}}=27+3 \times \frac{1}{3}=28 .
\end{array}
$$
It is verified that $m=1, m=2$ do not meet the requirements.
Hence $m \geqslant 3$.
$$
\begin{array}{l}
\text { By } \frac{a_{1}^{3}+a_{2}^{3}+a_{3}^{3}}{3} \geqslant a_{1} a_{2} a_{3}, \\
\frac{a_{1}^{3}+a_{2}^{3}+a_{4}^{3}}{3} \geqslant a_{1} a_{2} a_{4}, \\
\frac{a_{1}^{3}+a_{3}^{3}+a_{4}^{3}}{3} \geqslant a_{1} a_{3} a_{4}, \\
\frac{a_{2}^{3}+a_{3}^{3}+a_{4}^{3}}{3} \geqslant a_{2} a_{3} a_{4}, \\
a_{1} a_{2} a_{3} a_{4}=1, \\
\text { then } a_{1}^{3}+a_{2}^{3}+a_{3}^{3}+a_{4}^{3} \\
\geqslant a_{1} a_{2} a_{3}+a_{1} a_{2} a_{4}+a_{1} a_{3} a_{4}+a_{2} a_{3} a_{4} \\
=\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\frac{1}{a_{4}} .
\end{array}
$$
Therefore, $m=3$ meets the requirements:
Thus, the smallest positive integer $m$ is 3.
|
3
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
20. Let the ellipse be $\frac{x^{2}}{a^{2}}+y^{2}=1(a>1), \operatorname{Rt} \triangle A B C$ with $A(0,1)$ as the right-angle vertex, and sides $A B, B C$ intersecting the ellipse at points $B, C$. If the maximum area of $\triangle A B C$ is $\frac{27}{8}$, find the value of $a$.
|
20. Let $l_{A B}: y=k x+1(k>0)$. Then $l_{A C}: y=-\frac{1}{k} x+1$.
From $\left\{\begin{array}{l}y=k x+1, \\ \frac{x^{2}}{a^{2}}+y^{2}=1,\end{array}\right.$,
$\left(1+a^{2} k^{2}\right) x^{2}+2 a^{2} k x=0$
$\Rightarrow x_{B}=\frac{-2 a^{2} k}{1+a^{2} k^{2}}$.
Thus, $|A B|=\sqrt{1+k^{2}} \cdot \frac{2 a^{2} k}{1+a^{2} k^{2}}$.
Similarly, $|A C|=\sqrt{1+\frac{1}{k^{2}}} \cdot \frac{2 a^{2} k}{a^{2}+k^{2}}$.
Therefore, $S_{\triangle A B C}=\frac{1}{2}|A B||A C|$
$=2 a^{4} \frac{k\left(1+k^{2}\right)}{\left(1+a^{2} k^{2}\right)\left(a^{2}+k^{2}\right)}$
$=2 a^{4} \frac{k+\frac{1}{k}}{a^{2}\left(k^{2}+\frac{1}{k^{2}}\right)+a^{4}+1}$.
Let $t=k+\frac{1}{k} \geqslant 2$. Then
$S_{\triangle A B C}=\frac{2 a^{4} t}{a^{2} t^{2}+\left(a^{2}-1\right)^{2}}$
$=\frac{2 a^{4}}{a^{2} t+\frac{\left(a^{2}-1\right)^{2}}{t}}$.
And $a^{2} t+\frac{\left(a^{2}-1\right)^{2}}{t} \geqslant 2 a\left(a^{2}-1\right)$, when $t=\frac{a^{2}-1}{a}$, equality holds.
Thus, when $t=\frac{a^{2}-1}{a}$, $\left(S_{\triangle A B C}\right)_{\max }=\frac{a^{3}}{a^{2}-1}$.
From $\frac{a^{3}}{a^{2}-1}=\frac{27}{8}$
$\Rightarrow(a-3)\left(8 a^{2}-3 a-9\right)=0$
$\Rightarrow a=3, a=\frac{3+\sqrt{297}}{16}$.
From $\frac{a^{2}-1}{a}>2 \Rightarrow a>1+\sqrt{2}$.
Therefore, $a=\frac{3+\sqrt{297}}{16}$ is not valid, and is discarded.
Thus, $a=3$.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If the arithmetic mean of two positive numbers is $2 \sqrt{3}$, and the geometric mean is $\sqrt{3}$, what is the difference between these two numbers?
|
1. Let $x, y$ represent two numbers. Then
$$
\begin{array}{l}
\left\{\begin{array}{l}
x+y=4 \sqrt{3}, \\
x y=3
\end{array}\right. \\
\Leftrightarrow|x-y|=\sqrt{(x-y)^{2}} \\
=\sqrt{(x+y)^{2}-4 x y}=6,
\end{array}
$$
which means the difference between the two numbers is 6.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. For the quadratic function $y=x^{2}+b x+c$, the vertex of its graph is $D$, and it intersects the positive x-axis at points $A$ and $B$ from left to right, and the positive y-axis at point $C$. If $\triangle A B D$ and $\triangle O B C$ are both isosceles right triangles (where $O$ is the origin), then $b+2 c=$
|
2. 2 .
From the given information, we have
$$
\begin{array}{l}
C(0, c) 、 A\left(\frac{-\dot{b}-\sqrt{b^{2}-4 c}}{2}, 0\right), \\
B\left(\frac{-b+\sqrt{b^{2}-4 c}}{2}, 0\right), D\left(-\frac{b}{2},-\frac{b^{2}-4 c}{4}\right) .
\end{array}
$$
Draw $D E \perp A B$ at point $E$. Then $2 D E=A B$,
which means $2 \times \frac{b^{2}-4 c}{4}=\sqrt{b^{2}-4 c}$.
Solving this, we get $\sqrt{b^{2}-4 c}=0$ (discard) or 2 .
$$
\begin{array}{l}
\text { Also, } O C=O B \\
\Rightarrow c=\frac{-b+\sqrt{b^{2}-4 c}}{2} \\
\Rightarrow b+2 c=\sqrt{b^{2}-4 c}=2 .
\end{array}
$$
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The value of the positive integer $n$ that makes $2^{n}+256$ a perfect square is $\qquad$
|
3. 11.
When $n8$, $2^{n}+256=2^{8}\left(2^{n-8}+1\right)$, if it is a perfect square, then $2^{n-8}+1$ is the square of an odd number.
Let $2^{n-8}+1=(2 k+1)^{2}$ ( $k$ is a natural number). Then $2^{n-10}=k(k+1)$.
Since $k$ and $k+1$ are one odd and one even, then $k=1$, thus, $n=11$.
|
11
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 When $1 \leqslant x \leqslant 2$, simplify
$$
\sqrt{x+2 \sqrt{x-1}}+\sqrt{x-2 \sqrt{x-1}}
$$
The value equals $\qquad$ [5]
(2009, Beijing Middle School Mathematics Competition (Grade 8))
|
Solution 1
$$
p=\sqrt{x+2 \sqrt{x-1}}+\sqrt{x-2 \sqrt{x-1}} \text {. }
$$
Then $p^{2}=2 x+2 \sqrt{x^{2}-(2 \sqrt{x-1})^{2}}$
$$
=2 x+2 \sqrt{(x-2)^{2}} \text {. }
$$
Since $1 \leqslant x \leqslant 2$, therefore,
$$
\begin{array}{l}
p^{2}=2 x+2 \sqrt{(x-2)^{2}} \\
=2 x+2(2-x)=4 .
\end{array}
$$
And $p \geqslant 0$, so $p=2$.
Solution 2 Let
$$
p=\sqrt{x+2 \sqrt{x-1}}, q=\sqrt{x-2 \sqrt{x-1}}
$$
Then $p^{2}+q^{2}=2 x, p q=2-x$.
Since $(p+q)^{2}=p^{2}+q^{2}+2 p q$
$$
=2 x+2(2-x)=4(p \geqslant 0, q \geqslant 0) \text {, }
$$
Therefore, $p+q=2$, that is, the original expression $=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Let $f(x)$ be a continuous even function, and when $x>0$, $f(x)$ is a strictly monotonic function. Then the sum of all $x$ that satisfy $f(x) = f\left(\frac{x+3}{x+4}\right)$ is ( ).
(A) -3
(B) -8
(C) 3
(D) 8
|
8. B.
When $f(x)=f\left(\frac{x+3}{x+4}\right)$, i.e., $x=\frac{x+3}{x+4}$, we get $x^{2}+3 x-3=0$. At this point,
$$
x_{1}+x_{2}=-3 \text {. }
$$
Since $f(x)$ is a continuous even function, another scenario is $f(-x)=f\left(\frac{x+3}{x+4}\right)$, i.e., $-x=\frac{x+3}{x+4}$, which gives
$$
x^{2}+5 x+3=0 \text {. }
$$
Therefore, $x_{3}+x_{4}=-5$.
Thus, the sum of all $x$ is
$$
-3+(-5)=-8 \text {. }
$$
|
-8
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Let $a=\frac{\sqrt{5}-1}{2}$. Then $\frac{a^{5}+a^{4}-2 a^{3}-a^{2}-a+2}{a^{3}-a}=$ $\qquad$ [6] $(2008$, National Junior High School Mathematics Competition)
|
Solve: Given $a=\frac{\sqrt{5}-1}{2} \Rightarrow 2 a+1=\sqrt{5}$.
Square both sides and simplify to get
$$
a^{2}+a=1 \text {. }
$$
Therefore, the original expression is
$$
\begin{array}{l}
=\frac{a^{3}\left(a^{2}+a\right)-2 a^{3}-\left(a^{2}+a\right)+2}{a(a+1)(a-1)} \\
=\frac{1-a^{3}}{\left(a^{2}+a\right)(a-1)} \\
=-\left(a^{2}+a+1\right)=-2 .
\end{array}
$$
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 Given $a=\sqrt{5}-1$. Then the value of $2 a^{3}+7 a^{2}-2 a$ -12 is $\qquad$ [7]
(2010) "Mathematics Weekly" Cup National Junior High School Mathematics Competition)
|
Given $a=\sqrt{5}-1 \Rightarrow a+1=\sqrt{5}$. Squaring both sides, we get
$$
a^{2}+2 a=4 \text {. }
$$
Therefore, the original expression is
$$
\begin{array}{l}
=2 a\left(a^{2}+2 a\right)+3 a^{2}-2 a-12 \\
=3 a^{2}+6 a-12 \\
=3\left(a^{2}+2 a\right)-12=0 .
\end{array}
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Let $M=(5+\sqrt{24})^{2 n}\left(n \in \mathbf{N}_{+}\right), N$ be the fractional part of $M$. Then the value of $M(1-N)$ is $\qquad$
|
10.1.
Since $(5+\sqrt{24})^{2 n}+(5-\sqrt{24})^{2 n}$ is a positive integer, and $0<(5-\sqrt{24})^{2 n}<1$, therefore,
$$
\begin{array}{l}
N=1-(5-\sqrt{24})^{2 n} \\
M(1-N)=(5+\sqrt{24})^{2 n}(5-\sqrt{24})^{2 n}=1 .
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
16. As shown in Figure $2, P$ is a moving point on the parabola $y^{2}=2 x$, points $B$ and $C$ are on the $y$-axis, and the circle $(x-1)^{2}+y^{2}=1$ is inscribed in $\triangle P B C$. Find the minimum value of the area of $\triangle P B C$.
|
16. Let \( P\left(x_{0}, y_{0}\right) \), \( B(0, b) \), \( C(0, c) \), and assume \( b > c \). The line \( l_{P B} \) is given by \( y - b = \frac{y_{0} - b}{x_{0}} x \), which can be rewritten as
\[
\left(y_{0} - b\right) x - x_{0} y + x_{0} b = 0.
\]
The distance from the circle's center \((1, 0)\) to \( P B \) is 1, so
\[
\frac{\left|y_{0} - b + x_{0} b\right|}{\sqrt{\left(y_{0} - b\right)^{2} + x_{0}^{2}}} = 1.
\]
Thus,
\[
\left(y_{0} - b\right)^{2} + x_{0}^{2} = \left(y_{0} - b\right)^{2} + 2 x_{0} b \left(y_{0} - b\right) + x_{0}^{2} b^{2}.
\]
Given \( x_{0} > 2 \), the equation simplifies to
\[
\left(x_{0} - 2\right) b^{2} + 2 y_{0} b - x_{0} = 0.
\]
Similarly,
\[
\left(x_{0} - 2\right) c^{2} + 2 y_{0} c - x_{0} = 0.
\]
Therefore,
\[
b + c = \frac{-2 y_{0}}{x_{0} - 2}, \quad b c = \frac{-x_{0}}{x_{0} - 2}.
\]
Thus,
\[
(b - c)^{2} = \frac{4 x_{0}^{2} + 4 y_{0}^{2} - 8 x_{0}}{\left(x_{0} - 2\right)^{2}}.
\]
Since \( P\left(x_{0}, y_{0}\right) \) lies on the parabola, \( y_{0}^{2} = 2 x_{0} \). Therefore,
\[
(b - c)^{2} = \frac{4 x_{0}^{2}}{\left(x_{0} - 2\right)^{2}} \Rightarrow b - c = \frac{2 x_{0}}{x_{0} - 2}.
\]
Hence,
\[
S_{\triangle P B C} = \frac{1}{2} (b - c) x_{0} = \frac{x_{0}}{x_{0} - 2} \cdot x_{0} = \left(x_{0} - 2\right) + \frac{4}{x_{0} - 2} + 4 \geqslant 2 \sqrt{4} + 4 = 8.
\]
The equality holds when \( \left(x_{0} - 2\right)^{2} = 4 \), which gives \( x_{0} = 4 \) and \( y_{0} = \pm 2 \sqrt{2} \).
Therefore, the minimum value of \( S_{\triangle P B C} \) is 8.
(Provided by Shi Dongyang)
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $x$ be a positive integer, and $x<50$. Then the number of $x$ such that $x^{3}+11$ is divisible by 12 is $\qquad$.
|
$$
\begin{array}{l}
x^{3}+11=(x-1)\left(x^{2}+x+1\right)+12 \\
=(x-1) x(x+1)+(x-1)+12 .
\end{array}
$$
Notice that $6 \mid(x-1) x(x+1)$, so $6 \mid(x-1)$;
also, $2 \mid \left(x^{2}+x+1\right)$, hence $12 \mid (x-1)$.
Since $x<50$, we have
$$
x-1=0,12,24,36,48 \text {. }
$$
Thus, $x=1,13,25,37,49$, a total of 5.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. As shown in Figure 5, in $\triangle A B C$, $D$ is the midpoint of side $A B$, and $G$ is the midpoint of $C D$. A line through $G$ intersects $A C$ and $B C$ at points $P$ and $Q$, respectively. Then the value of $\frac{C A}{C P}+\frac{C B}{C Q}$ is $\qquad$.
|
4. 4 .
As shown in Figure 12, draw $A E / / P Q$ intersecting $C D$ at point $E$, and $B F / / P Q$ intersecting the extension of $C D$ at point $F$. Then $A E / / B F$.
Since $A D = B D$, therefore, $E D = F D$.
$$
\begin{array}{c}
\text { Also, } \frac{C A}{C P} = \frac{C E}{C G}, \\
\frac{C B}{C O} = \frac{C F}{C G}, \\
C E = C D - D E, C F = C D + F D, \\
\text { hence } \frac{C A}{C P} + \frac{C B}{C Q} = \frac{C E + C F}{C G} = \frac{2 C D}{C G} = 4 .
\end{array}
$$
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. If $n \in \mathbf{N}, n \geqslant 2, a_{i} \in\{0,1, \cdots, 9\}$
$$
\begin{array}{l}
(i=1,2, \cdots, n), a_{1} a_{2} \neq 0 \text {, and } \\
\sqrt{a_{1} a_{2} \cdots a_{n}}-\sqrt{a_{2} a_{3} \cdots a_{n}}=a_{1}, \\
\end{array}
$$
then $n=$ $\qquad$, where $\overline{a_{1} a_{2} \cdots a_{n}}$ is the $n$-digit number formed by $a_{1}, a_{2}, \cdots$ $a_{n}$.
|
2. 2 .
Let $x=\overline{a_{2} a_{3} \cdots a_{n}} \in \mathbf{N}_{+}$. Then $\overline{a_{1} a_{2} \cdots a_{n}}=10^{n-1} a_{1}+x$.
By the given condition, $\sqrt{10^{n-1} a_{1}+x}-\sqrt{x}=a_{1}$. Therefore, $10^{n-1} a_{1}+x=x+a_{1}^{2}+2 a_{1} \sqrt{x}$.
Thus, $10^{n-1}-2 \sqrt{x}=a_{1}$.
If $n \geqslant 3$, then
$$
\begin{array}{l}
10^{n-1}-2 \sqrt{x} \geqslant 10^{n-1}-2 \sqrt{10^{n-1}} \\
=\left(\sqrt{10^{n-1}}-1\right)^{2}-1 \\
\geqslant\left(\sqrt{10^{2}}-1\right)^{2}-1=80,
\end{array}
$$
which contradicts $a_{1} \leqslant 9$.
Therefore, $n=2$.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. On a straight line, three points $A$, $B$, and $C$ are arranged in sequence, and $A B=6, A C=24, D$ is a point outside the line, and $D A$ $\perp A B$. When $\angle B D C$ takes the maximum value, $A D=$ $\qquad$ .
|
5. 12 .
Let $\angle B D C=\theta\left(\theta<90^{\circ}\right), \triangle B C D$'s circumcircle $\odot O$ has a radius of $R$. Then $\sin \theta=\frac{B C}{2 R}$.
When $R$ decreases, $\theta$ increases.
Therefore, when $\odot O$ is tangent to $A D$ at point $D$, $\theta$ is maximized.
At this time, $A D^{2}=A B \cdot A C=12^{2} \Rightarrow A D=12$.
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given the set $M=\{2,0,11\}$. If $A \varsubsetneqq M$, and $A$ contains at least one even number, then the number of sets $A$ that satisfy the condition is $\qquad$ .
|
1. 5.
The sets $A$ that satisfy the conditions are $\{2\}, \{0\}, \{2,0\}, \{2,11\}, \{0,11\}$. There are 5 in total.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Let the function
$$
f(x)=\left\{\begin{array}{ll}
\frac{1}{p}, & x=\frac{q}{p} ; \\
0, & x \neq \frac{q}{p},
\end{array}\right.
$$
where $p$ and $q$ are coprime, and $p \geqslant 2$. Then the number of $x$ values that satisfy $x \in[0,1]$ and $f(x)>\frac{1}{5}$ is $\qquad$ .
|
7.5.
Obviously, $x=\frac{q}{p}$ (otherwise, $f(x)=0$). At this point, from $f(x)=\frac{1}{p}>\frac{1}{5}$, we get $p<5$, i.e., $p=2,3,4$. When $p=2$, $x=\frac{1}{2}$; when $p=3$, $x=\frac{1}{3}, \frac{2}{3}$; when $p=4$, $x=\frac{1}{4}, \frac{3}{4}$.
Therefore, there are 5 values of $x$ that satisfy the condition.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Find all odd prime numbers $p$ such that
$$
p \mid \sum_{k=1}^{2011} k^{p-1} .
$$
|
If $p>2011$, then for $k(1 \leqslant k \leqslant 2011)$ by Fermat's Little Theorem we have
$k^{p-1} \equiv 1(\bmod p)$.
Thus $\sum_{k=1}^{2011} k^{p-1} \equiv 2011 \not\equiv 0(\bmod p)$, a contradiction.
Therefore, $p \leqslant 2011$.
Let $2011=p q+r(0 \leqslant r < q)$, then if $r>q$, we have $q=r$, so
$$
2011=p q+r=(p+1) r \text {. }
$$
This implies $p$ is even, which contradicts $p$ being a prime.
If $p \leqslant q$, then $2011=p q+r \geqslant p^{2}$.
Thus, the odd primes satisfying $p^{2} \leqslant 2011$ are
$$
3,5,7,11,13, 17, 19,23,29,31,37,41,43 \text {. }
$$
Checking, we find that only $p=3$ satisfies the condition.
In summary, $p=3$.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The equation concerning $x, y$
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{x y}=\frac{1}{2011}
$$
has $\qquad$ groups of positive integer solutions $(x, y)$.
|
3. 12 .
From $\frac{1}{x}+\frac{1}{y}+\frac{1}{x y}=\frac{1}{2011}$, we get
$$
\begin{array}{l}
x y-2011 x-2011 y-2011=0 \\
\Rightarrow(x-2011)(y-2011) \\
\quad=2011 \times 2012 \\
=2^{2} \times 503 \times 2011 .
\end{array}
$$
Thus, the positive integer solutions of the original equation are
$$
(2+1)(1+1)(1+1)=12 \text { (sets). }
$$
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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